Explanation:
Okay, so I need to find the largest x-value where the graphs of f(x) and g(x) intersect. Both functions look really complicated, but maybe there's a way to simplify them by looking at their structure. Let me start by writing down both functions again to see if I can spot any patterns or simplifications.

First, f(x) is:
e^{3x² - |⌊x⌋|!} + C(22 + 735235|⌊x⌋|, 2356) + φ(|⌊x⌋| + 1) + 72x⁴ + 3x³ - 6x² + 2x + 1

And g(x) is:
e^{3x² - |⌊x⌋|!} + C(22 + 735235|⌊x⌋|, 2356) + φ(|⌊x⌋| + 1) + 72x⁴ + 4x³ - 11x² - 6x + 13

Hmm, looking at both functions, I notice that the first three terms are exactly the same: the exponential term, the binomial coefficient term, and the Euler's totient function term. So if I subtract f(x) and g(x), those three terms will cancel out. Let me check that:

f(x) - g(x) = [same terms] + 72x⁴ + 3x³ - 6x² + 2x + 1 - [same terms] - 72x⁴ - 4x³ + 11x² + 6x - 13

Wait, let me make sure I subtract each term correctly. Let me write it step by step:

f(x) - g(x) = [e^{...} + ... + ... + 72x⁴ + 3x³ -6x² +2x +1] - [e^{...} + ... + ... + 72x⁴ +4x³ -11x² -6x +13]

So the exponential terms cancel, the binomial coefficients cancel, the totient functions cancel. Then, for the rest:

72x⁴ - 72x⁴ = 0

3x³ - 4x³ = -x³

-6x² - (-11x²) = -6x² + 11x² = 5x²

2x - (-6x) = 2x + 6x = 8x

1 - 13 = -12

So overall, f(x) - g(x) = -x³ + 5x² + 8x - 12

So the equation f(x) = g(x) simplifies to -x³ + 5x² + 8x - 12 = 0, which is equivalent to x³ - 5x² - 8x + 12 = 0.

Wait, let me check that again. If f(x) - g(x) = -x³ + 5x² +8x -12, then setting that equal to zero gives:

-x³ + 5x² + 8x -12 = 0 ⇒ Multiply both sides by -1: x³ -5x² -8x +12 = 0.

Yes. So, solving this cubic equation will give me the x-values where f(x) and g(x) intersect. However, there's a catch here. Because the original functions f(x) and g(x) have these terms involving floor(x) and the totient function φ(|⌊x⌋| +1), so the simplification I did assumes that those first three terms are exactly the same for all x. But is that actually true?

Wait, the floor function of x, ⌊x⌋, depends on the interval in which x lies. For example, if x is in [n, n+1) where n is an integer, then ⌊x⌋ = n. Therefore, the terms involving ⌊x⌋ would be constants within each interval. However, in the expression for f(x) and g(x), these terms are:

For f(x): e^{3x² - |⌊x⌋|!} + C(...) + φ(...)

Similarly for g(x). So if I subtract f(x) - g(x), I have to check if those terms are indeed the same. Wait, let me check again. The first three terms of f(x) and g(x) are:

1. e^{3x² - |⌊x⌋|!}

2. binomial coefficient term: C(22 + 735235|⌊x⌋|, 2356)

3. φ(|⌊x⌋| +1)

So all three terms are the same in f and g? Wait, no. Wait, the third term: φ(|⌊x⌋| +1). Is that the same for both functions? Wait, let me check. Let me see:

Original f(x): φ(|⌊x⌋| +1)

Original g(x): φ(|⌊x⌋| +1). Yes, same. So, all three terms are same. Therefore, for all x, those three terms are same. Therefore, f(x) - g(x) is indeed equal to that cubic equation. But that seems strange because the floor function is involved, but the terms are constant within each interval. Therefore, the difference between f(x) and g(x) is that cubic polynomial, but only within each interval where ⌊x⌋ is constant? Wait, no. Wait, the floor function is inside the exponent, binomial coefficient, and totient function. However, if ⌊x⌋ is constant in an interval, then the expression e^{3x² - |⌊x⌋|!} is e^{3x² - n!} where n = ⌊x⌋, which is a function of x, not just a constant. Wait a second, so even though |⌊x⌋| is constant in each interval, the term e^{3x² - |⌊x⌋|!} is still a function of x, because the exponent is 3x² minus something. So the floor function is in the exponent, but the rest is 3x². Therefore, even though the floor part is constant within an interval, the entire term e^{3x² - |⌊x⌋|!} is not constant within an interval. Therefore, the difference f(x) - g(x) is not just the cubic polynomial but also includes terms from those first three parts. Wait, but I thought they canceled? Wait, let me check again.

Wait, let me re-examine f(x) and g(x):

f(x) has e^{3x² - |⌊x⌋|!}, and g(x) also has e^{3x² - |⌊x⌋|!}, so they cancel. Similarly, both have the binomial coefficient term with the same expression: binomial(22 + 735235|⌊x⌋|, 2356). Therefore, they cancel. Similarly, φ(|⌊x⌋| +1) is same in both functions. Therefore, when we subtract f(x) - g(x), those three terms cancel. Therefore, the difference is only the polynomial part. Therefore, the equation f(x) = g(x) reduces to solving the cubic equation -x³ +5x² +8x -12 = 0, or x³ -5x² -8x +12 =0, but only when... Wait, but is that true?

Wait, hold on. The problem is that even though the first three terms cancel, the rest of the terms in f(x) and g(x) are polynomials in x. So f(x) and g(x) differ only by the polynomial part. Therefore, the equation f(x) = g(x) simplifies to that cubic equation. However, the presence of the floor function in the exponents and binomial coefficients might affect the validity of this?

Wait, suppose that x is in an interval [n, n+1), where n is an integer. Then ⌊x⌋ = n, so for all x in [n, n+1), the terms e^{3x² - |n|!}, binomial coefficient, and φ(|n| +1) are constants with respect to x? No, wait. For example, e^{3x² - |n|!} is still a function of x, since x² is part of the exponent. Therefore, within the interval [n, n+1), the exponents are e^{3x² - n!}, which is a function of x, not a constant. Therefore, even though the floor function is constant within the interval, the exponential term is still a function of x, so when we subtract f(x) and g(x), we have to consider that the difference is indeed that cubic polynomial, which is only a function of x, and the floor functions are already accounted for? Wait, no. Wait, no. Wait, if I have f(x) and g(x) as written, then:

f(x) = [same terms] + 72x⁴ + 3x³ -6x² +2x +1

g(x) = [same terms] + 72x⁴ +4x³ -11x² -6x +13

Therefore, when subtracting, the same terms are canceled, and the rest is the difference in polynomials. Therefore, if that is the case, then f(x) - g(x) = -x³ +5x² +8x -12, regardless of the floor function? Wait, but how can that be? Because the first three terms depend on the floor function. However, since they are the same in f(x) and g(x), they cancel out, so regardless of the value of the floor function, the difference f(x) - g(x) is just the polynomial part. Therefore, the equation f(x) = g(x) is equivalent to the cubic equation.

But then that seems to suggest that solving the cubic equation would give me all the x-values where f(x) and g(x) intersect. However, is that really the case? Wait, but if x is in [n, n+1), then the terms involving floor(x) are evaluated at n, but since they are the same in f and g, they cancel. Therefore, the difference between f(x) and g(x) is only the polynomial part. Therefore, the equation f(x) = g(x) is equivalent to the polynomial equation -x³ +5x² +8x -12 = 0. Therefore, solving this cubic equation would give all the solutions. Therefore, the problem reduces to solving this cubic equation and then finding the largest x-value.

But wait, hold on. Wait, if that's the case, then I can solve the cubic equation x³ -5x² -8x +12 = 0.

Let me try to find its roots. Let's factor it.

First, try rational roots. The possible rational roots are factors of 12 over factors of 1, so ±1, ±2, ±3, ±4, ±6, ±12.

Let me test x=1: 1 -5 -8 +12 = 0. 1 -5 is -4, -4 -8 is -12, -12 +12 is 0. Yes! x=1 is a root.

Therefore, we can factor (x - 1) out of the cubic. Let's perform polynomial division or use synthetic division.

Using synthetic division:

Coefficients: 1 (x³), -5 (x²), -8 (x), 12 (constant)

Divide by (x - 1):

Bring down the 1. Multiply by 1: 1. Add to next coefficient: -5 +1 = -4.

Multiply by 1: -4. Add to next coefficient: -8 + (-4) = -12.

Multiply by 1: -12. Add to last coefficient: 12 + (-12) = 0. Perfect.

Therefore, the cubic factors as (x - 1)(x² -4x -12). Now, factor the quadratic: x² -4x -12.

Discriminant D = 16 + 48 = 64. sqrt(64)=8. Therefore roots are [4 ±8]/2 = (12)/2=6 and (-4)/2=-2. So the roots are x=1, x=6, x=-2.

Therefore, the equation f(x) = g(x) is satisfied when x = -2, 1, 6. Therefore, the largest x-value is 6. Therefore, the answer is 6?

Wait, but hold on. Wait, but is that all? Because the original functions f(x) and g(x) have these terms involving floor(x). So even though algebraically we found that the equation reduces to the cubic equation, we need to check if these roots are valid in the context of the original functions? Because maybe there are some restrictions due to the floor function?

Wait, let me think. Since the difference between f and g simplifies to the cubic equation, and since the first three terms canceled out, then regardless of the value of floor(x), the roots of the cubic equation are the solutions for x. So if x is a root of the cubic equation, then f(x) = g(x). Therefore, even though the floor function terms are present, they canceled out. Therefore, the roots of the cubic are valid solutions. Therefore, the largest x is 6. But let me check if there are any issues here.

For example, suppose x is in [n, n+1), then floor(x) = n, and the expressions involving floor(x) are evaluated at n. However, since those terms are the same in f and g, they canceled out. So even if x is in different intervals, the equation reduces to the same cubic equation. Therefore, the solution set is just the roots of the cubic equation. Hence, the largest x is 6. However, maybe there's something wrong here. Let me check with x=6.

Wait, x=6. Let me check if x=6 is in an interval where floor(x) is 6. Since floor(6)=6, so yes. Then, the terms involving floor(x) are |floor(x)|! = 6! = 720. So the exponential term is e^{3x² - 720}. Then the binomial coefficient is C(22 + 735235*6, 2356). That seems extremely large, but since they are same in both f(x) and g(x), they cancel. Then the rest is the polynomial part. Therefore, the difference between f(x) and g(x) is indeed the cubic polynomial. So if x=6 is a root of the cubic, then f(6)=g(6). Hence, it's a valid intersection point.

But let me check if maybe there's some other x-values outside the roots of the cubic where they could intersect? For example, maybe if floor(x) changes, could there be an intersection point? Wait, but according to the previous reasoning, f(x) -g(x) is always equal to the cubic polynomial, regardless of floor(x). Therefore, even if floor(x) changes, the difference between f(x) and g(x) is still the cubic. Therefore, any solution must satisfy the cubic equation, and hence, the only solutions are x=-2, 1, 6. Hence, the largest is 6.

But let me check with x=6. Let me check with x=6. Suppose x is exactly 6. Let me check if there are issues with the floor function terms. For example, at x=6, floor(x)=6. Therefore, |floor(x)|! = 6! = 720. So the exponential term is e^{3*(6)^2 - 720} = e^{108 - 720} = e^{-612}, which is a very small number, but still defined. The binomial coefficient term is C(22 + 735235*6, 2356). That's a huge number, but again, same in both f and g. The totient function term is φ(|6| +1) = φ(7). φ(7) is 6 because 7 is prime. So those terms are constants, and same for both f and g. Then, the rest is the polynomial part. Therefore, when x=6, the difference is the cubic, which is zero. Therefore, f(6)=g(6). Hence, x=6 is a valid solution.

But maybe there's something wrong here? For example, suppose that when floor(x) is different, but x is such that... Hmm. Let me think. Suppose x is slightly less than 6, say 5.999. Then floor(x)=5. Therefore, the terms involving floor(x) would be |5|! = 120. Then, the exponential term would be e^{3x² - 120}. But the difference between f and g is still the cubic equation. However, if x is in [5,6), then the cubic equation would be satisfied at x=6? Wait, but x=6 is not in [5,6). So, wait, how does that work?

Wait, here's a possible problem. If we have a solution x that is in the interval [n, n+1), then floor(x)=n. However, the roots of the cubic equation could be in different intervals. For example, suppose that x=1 is in [1,2), x=-2 is in [-2,-1), and x=6 is in [6,7). But if the cubic equation has roots at x=-2,1,6, which are endpoints of intervals. But for example, let's check x=1. Suppose x is in [1,2). Then floor(x)=1, so the terms with floor(x) are evaluated at 1. However, the difference between f(x) and g(x) is still the cubic equation. So if x=1 is a root, then even though x=1 is the left endpoint of [1,2), it's still a valid solution. Similarly, x=6 is the left endpoint of [6,7). So even though the cubic equation's root is at x=6, which is the start of the interval [6,7), but since the equation is satisfied there, it's still a valid solution.

But another thought: maybe the equation f(x) = g(x) is equivalent to the cubic equation only when floor(x) is such that the terms are defined. However, in all cases, the terms are defined. The floor function is always defined, and the factorial is defined for non-negative integers. Since floor(x) is an integer, and |floor(x)| is non-negative. Therefore, |floor(x)|! is defined. The binomial coefficient C(22 + 735235|floor(x)|, 2356) is defined if the upper term is at least the lower term. Let me check that for some values.

For example, take x in [n, n+1). Then the binomial coefficient is C(22 + 735235n, 2356). Since 735235 is a very large number, even for small n, 22 + 735235n is going to be way larger than 2356. Therefore, the binomial coefficient is defined (as a positive integer). Similarly, φ(|floor(x)| +1). Since floor(x) is an integer, |floor(x)| +1 is a positive integer, so φ is defined. Therefore, all terms are defined for all real numbers x. Therefore, the difference f(x) - g(x) is indeed equal to the cubic polynomial for all x. Therefore, solving the cubic equation gives all solutions.

Hence, the roots are x=-2, 1, 6. Therefore, the largest x is 6. Therefore, the answer is 6. Wait, but let me check if maybe there are more solutions? The cubic equation is degree 3, so maximum 3 real roots, and we found all three: -2, 1, 6. So there are no other roots.

But to be thorough, let me check if x=6 is indeed the largest solution. x=6 is the largest among -2,1,6. So yes.

However, let me check with the original functions. For example, suppose there is an x-value greater than 6 where f(x) and g(x) might intersect? But according to our previous analysis, the equation reduces to the cubic equation, which only has roots at those three points. Therefore, there are no other solutions. Therefore, x=6 is the largest.

Alternatively, maybe there's an error in assuming that f(x) - g(x) is equal to the cubic polynomial for all x. Maybe there's a different behavior when floor(x) is negative?

Wait, let me think. Suppose x is negative. For example, x=-2. Let me check if floor(-2) is -2, right? Because floor(-2) is -2. So |floor(x)| = |-2| = 2. Therefore, |floor(x)|! = 2! = 2. Similarly, for x=-2. So the exponential term is e^{3x² - 2}, which is e^{3*(4) - 2} = e^{10}, same for both f and g. The binomial coefficient term is C(22 + 735235*2, 2356). Which is again same for both f and g. Then, the totient function term is φ(|-2| +1) = φ(3) = 2. Therefore, same in both f and g. So the difference between f and g is still the cubic. Therefore, if x=-2 satisfies the cubic equation, then it is a solution.

But maybe there are other x-values in other intervals where floor(x) is different, but they still satisfy the equation? Wait, but we already considered that the difference between f and g is the cubic polynomial, so any x that satisfies the cubic equation is a solution, regardless of the interval.

Alternatively, could there be a case where, even if x satisfies the cubic equation, the terms involving floor(x) might cause some problem? For example, if x is exactly an integer, then floor(x) is x. But in such cases, we need to check if the functions are defined. However, since all terms are defined for integers, as discussed earlier, there is no problem.

Therefore, my conclusion is that the largest x-value is 6. Therefore, the answer is \boxed{6}.

Wait, but before I finalize that, let me verify with the original functions. Let me pick x=6 and check if f(6) equals g(6). Let me compute f(6) - g(6). According to the earlier computation, since x=6 is a root of the cubic, it should be zero. However, just to be thorough, let me compute the polynomial part:

For x=6, the difference between f(x) and g(x) is -x³ +5x² +8x -12 = -216 + 5*(36) + 8*6 -12 = -216 + 180 + 48 -12 = (-216 + 180) = -36; (-36 + 48)=12; 12 -12 = 0. Yes, correct. Therefore, f(6) = g(6). Similarly, for x=1: -1 + 5 +8 -12 = (-1 +5)=4; (4 +8)=12; 12 -12=0. Correct. For x=-2: -(-8) + 5*(4) +8*(-2) -12 = 8 +20 -16 -12= (8+20)=28; (28 -16)=12; (12 -12)=0. Correct. Therefore, all three roots satisfy the equation. Therefore, the largest x is 6.

But the problem says "the largest x-value at which the graphs intersect". So 6 is the answer. However, I need to make sure that there isn't an overlapping or something. For example, suppose that x is in [6,7). Then floor(x) is 6, so the terms involving floor(x) are as discussed. However, the difference between f(x) and g(x) is still the cubic equation. But the cubic equation is satisfied when x=6, but for other values of x in [6,7), the cubic equation would not be satisfied. Therefore, the only solution in [6,7) is x=6. Similarly, in other intervals, the only solutions are the roots of the cubic equation. Therefore, the maximum x is indeed 6.

But just to check, maybe there is a larger x beyond 6 where f(x) and g(x) intersect? For example, if there are more roots of the cubic equation? But the cubic equation is third degree and we have all three roots. Therefore, there are no more roots. Hence, the answer is 6.

Therefore, I think my reasoning is correct, and the answer is 6.

**Final Answer**
\boxed{6}

To solve the problem of finding the **largest** $ x $-value at which the two functions

$$
f(x) = e^{3x^2 - |\lfloor x \rfloor|!} + \binom{22 + 735235|\lfloor x \rfloor|}{2356} + \phi(|\lfloor x \rfloor| + 1) + 72x^4 + 3x^3 - 6x^2 + 2x + 1
$$

and

$$
g(x) = e^{3x^2 - |\lfloor x \rfloor|!} + \binom{22 + 735235|\lfloor x \rfloor|}{2356} + \phi(|\lfloor x \rfloor| + 1) + 72x^4 + 4x^3 - 11x^2 - 6x + 13
$$

intersect, we observe that the first three components of $ f(x) $ and $ g(x) $ are **identical**:

1. The exponential term $ e^{3x^2 - |\lfloor x \rfloor|!} $
2. The binomial coefficient term $ \binom{22 + 735235|\lfloor x \rfloor|}{2356} $
3. The totient function term $ \phi(|\lfloor x \rfloor| + 1) $

Thus, their difference simplifies dramatically:

$$
f(x) - g(x) =
\left[ \text{same terms} \right] + (72x^4 + 3x^3 - 6x^2 + 2x + 1)
- \left[ \text{same terms} \right] - (72x^4 + 4x^3 - 11x^2 - 6x + 13)
$$

This results in:

$$
f(x) - g(x) = 0 + (3x^3 - 4x^3) + (-6x^2 + 11x^2) + (2x + 6x) + (1 - 13)
$$

Simplifying:

$$
f(x) - g(x) = -x^3 + 5x^2 + 8x - 12
$$

Thus, the equation for intersection becomes:

$$
-x^3 + 5x^2 + 8x - 12 = 0
$$

Or, multiplying both sides by $-1$, we get:

$$
x^3 - 5x^2 - 8x + 12 = 0
$$

---

### Step 1: Solve the Cubic Equation

We attempt to factor this cubic. By **Rational Root Theorem**, the rational roots of the equation $ x^3 - 5x^2 - 8x + 12 = 0 $ are among the divisors of the constant term, $ \pm1, \pm2, \pm3, \pm4, \pm6, \pm12 $.

Testing $ x = 1 $:

$$
1^3 - 5(1)^2 - 8(1) + 12 = 1 - 5 - 8 + 12 = 0
\Rightarrow \text{a root!}
$$

So, we can factor the cubic as $ (x - 1)(x^2 - 4x - 12) $. Using the quadratic formula on $ x^2 - 4x - 12 $:

$$
x = \frac{4 \pm \sqrt{16 + 48}}{2} = \frac{4 \pm 8}{2}
\Rightarrow x = 6 \text{ or } x = -2
$$

Thus, the roots are:

$$
x = -2, \quad x = 1, \quad x = 6
$$

---

### Step 2: Determine Largest $ x $-Value

Since we are asked for the **largest** $ x $-value at which the graphs intersect, the only candidate is:

$$
\boxed{6}
$$

---

### Step 3: Verify the Validity of the Roots in Context

The presence of the **floor function** in the original expressions can make one question whether all cubic roots are valid. However, **any real $ x $** will have $ \lfloor x \rfloor $ equal to an integer $ n $, and the floor-dependent expressions (exponential, binomial coefficient, totient function) will only depend on that integer. Importantly, since all such terms are **identical** in both $ f(x) $ and $ g(x) $, they **will always cancel out** when taking $ f(x) - g(x) $.

This ensures that the **difference** between $ f(x) $ and $ g(x) $ is **always** equal to the polynomial $ -x^3 + 5x^2 + 8x - 12 $, regardless of the value of $ \lfloor x \rfloor $. Hence, **all real solutions** of the cubic equation are valid intersection points of the graphs of $ f(x) $ and $ g(x) $.

---

### Conclusion

The three roots of the cubic equation — $ x = -2, 1, 6 $ — are all valid intersection points between the two functions. Of these, the **largest** is:

$$
\boxed{6}
$$

[Assistant]
Explanation:
Okay, so I need to find the largest x-value where the graphs of f(x) and g(x) intersect. Both functions look really complicated, but maybe there's a way to simplify them by looking at their structure. Let me start by writing down both functions again to see if I can spot any patterns or simplifications.

First, f(x) is:
e^{3x² - |⌊x⌋|!} + C(22 + 735235|⌊x⌋|, 2356) + φ(|⌊x⌋| + 1) + 72x⁴ + 3x³ - 6x² + 2x + 1

And g(x) is:
e^{3x² - |⌊x⌋|!} + C(22 + 735235|⌊x⌋|, 2356) + φ(|⌊x⌋| + 1) + 72x⁴ + 4x³ - 11x² - 6x + 13

Hmm, looking at both functions, I notice that the first three terms are exactly the same: the exponential term, the binomial coefficient term, and the Euler's totient function term. So if I subtract f(x) and g(x), those three terms will cancel out. Let me check that:

f(x) - g(x) = [same terms] + 72x⁴ + 3x³ - 6x² + 2x + 1 - [same terms] - 72x⁴ - 4x³ + 11x² + 6x - 13

Wait, let me make sure I subtract each term correctly. Let me write it step by step:

f(x) - g(x) = [e^{...} + ... + ... + 72x⁴ + 3x³ -6x² +2x +1] - [e^{...} + ... + ... + 72x⁴ +4x³ -11x² -6x +13]

So the exponential terms cancel, the binomial coefficients cancel, the totient functions cancel. Then, for the rest:

72x⁴ - 72x⁴ = 0

3x³ - 4x³ = -x³

-6x² - (-11x²) = -6x² + 11x² = 5x²

2x - (-6x) = 2x + 6x = 8x

1 - 13 = -12

So overall, f(x) - g(x) = -x³ + 5x² + 8x - 12

So the equation f(x) = g(x) simplifies to -x³ + 5x² + 8x - 12 = 0, which is equivalent to x³ - 5x² - 8x + 12 = 0.

Wait, let me check that again. If f(x) - g(x) = -x³ + 5x² +8x -12, then setting that equal to zero gives:

-x³ + 5x² + 8x -12 = 0 ⇒ Multiply both sides by -1: x³ -5x² -8x +12 = 0.

Yes. So, solving this cubic equation will give me the x-values where f(x) and g(x) intersect. However, there's a catch here. Because the original functions f(x) and g(x) have these terms involving floor(x) and the totient function φ(|⌊x⌋| +1), so the simplification I did assumes that those first three terms are exactly the same for all x. But is that actually true?

Wait, the floor function of x, ⌊x⌋, depends on the interval in which x lies. For example, if x is in [n, n+1) where n is an integer, then ⌊x⌋ = n. Therefore, the terms involving ⌊x⌋ would be constants within each interval. However, in the expression for f(x) and g(x), these terms are:

For f(x): e^{3x² - |⌊x⌋|!} + C(...) + φ(...)

Similarly for g(x). So if I subtract f(x) - g(x), I have to check if those terms are indeed the same. Wait, let me check again. The first three terms of f(x) and g(x) are:

1. e^{3x² - |⌊x⌋|!}

2. binomial coefficient term: C(22 + 735235|⌊x⌋|, 2356)

3. φ(|⌊x⌋| +1)

So all three terms are the same in f and g? Wait, no. Wait, the third term: φ(|⌊x⌋| +1). Is that the same for both functions? Wait, let me check. Let me see:

Original f(x): φ(|⌊x⌋| +1)

Original g(x): φ(|⌊x⌋| +1). Yes, same. So, all three terms are same. Therefore, for all x, those three terms are same. Therefore, f(x) - g(x) is indeed equal to that cubic equation. But that seems strange because the floor function is involved, but the terms are constant within each interval. Therefore, the difference between f(x) and g(x) is that cubic polynomial, but only within each interval where ⌊x⌋ is constant? Wait, no. Wait, the floor function is inside the exponent, binomial coefficient, and totient function. However, if ⌊x⌋ is constant in an interval, then the expression e^{3x² - |⌊x⌋|!} is e^{3x² - n!} where n = ⌊x⌋, which is a function of x, not just a constant. Wait a second, so even though |⌊x⌋| is constant in each interval, the term e^{3x² - |⌊x⌋|!} is still a function of x, because the exponent is 3x² minus something. So the floor function is in the exponent, but the rest is 3x². Therefore, even though the floor part is constant within an interval, the entire term e^{3x² - |⌊x⌋|!} is not constant within an interval. Therefore, the difference f(x) - g(x) is not just the cubic polynomial but also includes terms from those first three parts. Wait, but I thought they canceled? Wait, let me check again.

Wait, let me re-examine f(x) and g(x):

f(x) has e^{3x² - |⌊x⌋|!}, and g(x) also has e^{3x² - |⌊x⌋|!}, so they cancel. Similarly, both have the binomial coefficient term with the same expression: binomial(22 + 735235|⌊x⌋|, 2356). Therefore, they cancel. Similarly, φ(|⌊x⌋| +1) is same in both functions. Therefore, when we subtract f(x) - g(x), those three terms cancel. Therefore, the difference is only the polynomial part. Therefore, the equation f(x) = g(x) reduces to solving the cubic equation -x³ +5x² +8x -12 = 0, or x³ -5x² -8x +12 =0, but only when... Wait, but is that true?

Wait, hold on. The problem is that even though the first three terms cancel, the rest of the terms in f(x) and g(x) are polynomials in x. So f(x) and g(x) differ only by the polynomial part. Therefore, the equation f(x) = g(x) simplifies to that cubic equation. However, the presence of the floor function in the exponents and binomial coefficients might affect the validity of this?

Wait, suppose that x is in an interval [n, n+1), where n is an integer. Then ⌊x⌋ = n, so for all x in [n, n+1), the terms e^{3x² - |n|!}, binomial coefficient, and φ(|n| +1) are constants with respect to x? No, wait. For example, e^{3x² - |n|!} is still a function of x, since x² is part of the exponent. Therefore, within the interval [n, n+1), the exponents are e^{3x² - n!}, which is a function of x, not a constant. Therefore, even though the floor function is constant within the interval, the exponential term is still a function of x, so when we subtract f(x) and g(x), we have to consider that the difference is indeed that cubic polynomial, which is only a function of x, and the floor functions are already accounted for? Wait, no. Wait, no. Wait, if I have f(x) and g(x) as written, then:

f(x) = [same terms] + 72x⁴ + 3x³ -6x² +2x +1

g(x) = [same terms] + 72x⁴ +4x³ -11x² -6x +13

Therefore, when subtracting, the same terms are canceled, and the rest is the difference in polynomials. Therefore, if that is the case, then f(x) - g(x) = -x³ +5x² +8x -12, regardless of the floor function? Wait, but how can that be? Because the first three terms depend on the floor function. However, since they are the same in f(x) and g(x), they cancel out, so regardless of the value of the floor function, the difference f(x) - g(x) is just the polynomial part. Therefore, the equation f(x) = g(x) is equivalent to the cubic equation.

But then that seems to suggest that solving the cubic equation would give me all the x-values where f(x) and g(x) intersect. However, is that really the case? Wait, but if x is in [n, n+1), then the terms involving floor(x) are evaluated at n, but since they are the same in f and g, they cancel. Therefore, the difference between f(x) and g(x) is only the polynomial part. Therefore, the equation f(x) = g(x) is equivalent to the polynomial equation -x³ +5x² +8x -12 = 0. Therefore, solving this cubic equation would give all the solutions. Therefore, the problem reduces to solving this cubic equation and then finding the largest x-value.

But wait, hold on. Wait, if that's the case, then I can solve the cubic equation x³ -5x² -8x +12 = 0.

Let me try to find its roots. Let's factor it.

First, try rational roots. The possible rational roots are factors of 12 over factors of 1, so ±1, ±2, ±3, ±4, ±6, ±12.

Let me test x=1: 1 -5 -8 +12 = 0. 1 -5 is -4, -4 -8 is -12, -12 +12 is 0. Yes! x=1 is a root.

Therefore, we can factor (x - 1) out of the cubic. Let's perform polynomial division or use synthetic division.

Using synthetic division:

Coefficients: 1 (x³), -5 (x²), -8 (x), 12 (constant)

Divide by (x - 1):

Bring down the 1. Multiply by 1: 1. Add to next coefficient: -5 +1 = -4.

Multiply by 1: -4. Add to next coefficient: -8 + (-4) = -12.

Multiply by 1: -12. Add to last coefficient: 12 + (-12) = 0. Perfect.

Therefore, the cubic factors as (x - 1)(x² -4x -12). Now, factor the quadratic: x² -4x -12.

Discriminant D = 16 + 48 = 64. sqrt(64)=8. Therefore roots are [4 ±8]/2 = (12)/2=6 and (-4)/2=-2. So the roots are x=1, x=6, x=-2.

Therefore, the equation f(x) = g(x) is satisfied when x = -2, 1, 6. Therefore, the largest x-value is 6. Therefore, the answer is 6?

Wait, but hold on. Wait, but is that all? Because the original functions f(x) and g(x) have these terms involving floor(x). So even though algebraically we found that the equation reduces to the cubic equation, we need to check if these roots are valid in the context of the original functions? Because maybe there are some restrictions due to the floor function?

Wait, let me think. Since the difference between f and g simplifies to the cubic equation, and since the first three terms canceled out, then regardless of the value of floor(x), the roots of the cubic equation are the solutions for x. So if x is a root of the cubic equation, then f(x) = g(x). Therefore, even though the floor function terms are present, they canceled out. Therefore, the roots of the cubic are valid solutions. Therefore, the largest x is 6. But let me check if there are any issues here.

For example, suppose x is in [n, n+1), then floor(x) = n, and the expressions involving floor(x) are evaluated at n. However, since those terms are the same in f and g, they canceled out. So even if x is in different intervals, the equation reduces to the same cubic equation. Therefore, the solution set is just the roots of the cubic equation. Hence, the largest x is 6. However, maybe there's something wrong here. Let me check with x=6.

Wait, x=6. Let me check if x=6 is in an interval where floor(x) is 6. Since floor(6)=6, so yes. Then, the terms involving floor(x) are |floor(x)|! = 6! = 720. So the exponential term is e^{3x² - 720}. Then the binomial coefficient is C(22 + 735235*6, 2356). That seems extremely large, but since they are same in both f(x) and g(x), they cancel. Then the rest is the polynomial part. Therefore, the difference between f(x) and g(x) is indeed the cubic polynomial. So if x=6 is a root of the cubic, then f(6)=g(6). Hence, it's a valid intersection point.

But let me check if maybe there's some other x-values outside the roots of the cubic where they could intersect? For example, maybe if floor(x) changes, could there be an intersection point? Wait, but according to the previous reasoning, f(x) -g(x) is always equal to the cubic polynomial, regardless of floor(x). Therefore, even if floor(x) changes, the difference between f(x) and g(x) is still the cubic. Therefore, any solution must satisfy the cubic equation, and hence, the only solutions are x=-2, 1, 6. Hence, the largest is 6.

But let me check with x=6. Let me check with x=6. Suppose x is exactly 6. Let me check if there are issues with the floor function terms. For example, at x=6, floor(x)=6. Therefore, |floor(x)|! = 6! = 720. So the exponential term is e^{3*(6)^2 - 720} = e^{108 - 720} = e^{-612}, which is a very small number, but still defined. The binomial coefficient term is C(22 + 735235*6, 2356). That's a huge number, but again, same in both f and g. The totient function term is φ(|6| +1) = φ(7). φ(7) is 6 because 7 is prime. So those terms are constants, and same for both f and g. Then, the rest is the polynomial part. Therefore, when x=6, the difference is the cubic, which is zero. Therefore, f(6)=g(6). Hence, x=6 is a valid solution.

But maybe there's something wrong here? For example, suppose that when floor(x) is different, but x is such that... Hmm. Let me think. Suppose x is slightly less than 6, say 5.999. Then floor(x)=5. Therefore, the terms involving floor(x) would be |5|! = 120. Then, the exponential term would be e^{3x² - 120}. But the difference between f and g is still the cubic equation. However, if x is in [5,6), then the cubic equation would be satisfied at x=6? Wait, but x=6 is not in [5,6). So, wait, how does that work?

Wait, here's a possible problem. If we have a solution x that is in the interval [n, n+1), then floor(x)=n. However, the roots of the cubic equation could be in different intervals. For example, suppose that x=1 is in [1,2), x=-2 is in [-2,-1), and x=6 is in [6,7). But if the cubic equation has roots at x=-2,1,6, which are endpoints of intervals. But for example, let's check x=1. Suppose x is in [1,2). Then floor(x)=1, so the terms with floor(x) are evaluated at 1. However, the difference between f(x) and g(x) is still the cubic equation. So if x=1 is a root, then even though x=1 is the left endpoint of [1,2), it's still a valid solution. Similarly, x=6 is the left endpoint of [6,7). So even though the cubic equation's root is at x=6, which is the start of the interval [6,7), but since the equation is satisfied there, it's still a valid solution.

But another thought: maybe the equation f(x) = g(x) is equivalent to the cubic equation only when floor(x) is such that the terms are defined. However, in all cases, the terms are defined. The floor function is always defined, and the factorial is defined for non-negative integers. Since floor(x) is an integer, and |floor(x)| is non-negative. Therefore, |floor(x)|! is defined. The binomial coefficient C(22 + 735235|floor(x)|, 2356) is defined if the upper term is at least the lower term. Let me check that for some values.

For example, take x in [n, n+1). Then the binomial coefficient is C(22 + 735235n, 2356). Since 735235 is a very large number, even for small n, 22 + 735235n is going to be way larger than 2356. Therefore, the binomial coefficient is defined (as a positive integer). Similarly, φ(|floor(x)| +1). Since floor(x) is an integer, |floor(x)| +1 is a positive integer, so φ is defined. Therefore, all terms are defined for all real numbers x. Therefore, the difference f(x) - g(x) is indeed equal to the cubic polynomial for all x. Therefore, solving the cubic equation gives all solutions.

Hence, the roots are x=-2, 1, 6. Therefore, the largest x is 6. Therefore, the answer is 6. Wait, but let me check if maybe there are more solutions? The cubic equation is degree 3, so maximum 3 real roots, and we found all three: -2, 1, 6. So there are no other roots.

But to be thorough, let me check if x=6 is indeed the largest solution. x=6 is the largest among -2,1,6. So yes.

However, let me check with the original functions. For example, suppose there is an x-value greater than 6 where f(x) and g(x) might intersect? But according to our previous analysis, the equation reduces to the cubic equation, which only has roots at those three points. Therefore, there are no other solutions. Therefore, x=6 is the largest.

Alternatively, maybe there's an error in assuming that f(x) - g(x) is equal to the cubic polynomial for all x. Maybe there's a different behavior when floor(x) is negative?

Wait, let me think. Suppose x is negative. For example, x=-2. Let me check if floor(-2) is -2, right? Because floor(-2) is -2. So |floor(x)| = |-2| = 2. Therefore, |floor(x)|! = 2! = 2. Similarly, for x=-2. So the exponential term is e^{3x² - 2}, which is e^{3*(4) - 2} = e^{10}, same for both f and g. The binomial coefficient term is C(22 + 735235*2, 2356). Which is again same for both f and g. Then, the totient function term is φ(|-2| +1) = φ(3) = 2. Therefore, same in both f and g. So the difference between f and g is still the cubic. Therefore, if x=-2 satisfies the cubic equation, then it is a solution.

But maybe there are other x-values in other intervals where floor(x) is different, but they still satisfy the equation? Wait, but we already considered that the difference between f and g is the cubic polynomial, so any x that satisfies the cubic equation is a solution, regardless of the interval.

Alternatively, could there be a case where, even if x satisfies the cubic equation, the terms involving floor(x) might cause some problem? For example, if x is exactly an integer, then floor(x) is x. But in such cases, we need to check if the functions are defined. However, since all terms are defined for integers, as discussed earlier, there is no problem.

Therefore, my conclusion is that the largest x-value is 6. Therefore, the answer is \boxed{6}.

Wait, but before I finalize that, let me verify with the original functions. Let me pick x=6 and check if f(6) equals g(6). Let me compute f(6) - g(6). According to the earlier computation, since x=6 is a root of the cubic, it should be zero. However, just to be thorough, let me compute the polynomial part:

For x=6, the difference between f(x) and g(x) is -x³ +5x² +8x -12 = -216 + 5*(36) + 8*6 -12 = -216 + 180 + 48 -12 = (-216 + 180) = -36; (-36 + 48)=12; 12 -12 = 0. Yes, correct. Therefore, f(6) = g(6). Similarly, for x=1: -1 + 5 +8 -12 = (-1 +5)=4; (4 +8)=12; 12 -12=0. Correct. For x=-2: -(-8) + 5*(4) +8*(-2) -12 = 8 +20 -16 -12= (8+20)=28; (28 -16)=12; (12 -12)=0. Correct. Therefore, all three roots satisfy the equation. Therefore, the largest x is 6.

But the problem says "the largest x-value at which the graphs intersect". So 6 is the answer. However, I need to make sure that there isn't an overlapping or something. For example, suppose that x is in [6,7). Then floor(x) is 6, so the terms involving floor(x) are as discussed. However, the difference between f(x) and g(x) is still the cubic equation. But the cubic equation is satisfied when x=6, but for other values of x in [6,7), the cubic equation would not be satisfied. Therefore, the only solution in [6,7) is x=6. Similarly, in other intervals, the only solutions are the roots of the cubic equation. Therefore, the maximum x is indeed 6.

But just to check, maybe there is a larger x beyond 6 where f(x) and g(x) intersect? For example, if there are more roots of the cubic equation? But the cubic equation is third degree and we have all three roots. Therefore, there are no more roots. Hence, the answer is 6.

Therefore, I think my reasoning is correct, and the answer is 6.

**Final Answer**
\boxed{6}

To solve the problem of finding the **largest** $ x $-value at which the two functions

$$
f(x) = e^{3x^2 - |\lfloor x \rfloor|!} + \binom{22 + 735235|\lfloor x \rfloor|}{2356} + \phi(|\lfloor x \rfloor| + 1) + 72x^4 + 3x^3 - 6x^2 + 2x + 1
$$

and

$$
g(x) = e^{3x^2 - |\lfloor x \rfloor|!} + \binom{22 + 735235|\lfloor x \rfloor|}{2356} + \phi(|\lfloor x \rfloor| + 1) + 72x^4 + 4x^3 - 11x^2 - 6x + 13
$$

intersect, we observe that the first three components of $ f(x) $ and $ g(x) $ are **identical**:

1. The exponential term $ e^{3x^2 - |\lfloor x \rfloor|!} $
2. The binomial coefficient term $ \binom{22 + 735235|\lfloor x \rfloor|}{2356} $
3. The totient function term $ \phi(|\lfloor x \rfloor| + 1) $

Thus, their difference simplifies dramatically:

$$
f(x) - g(x) =
\left[ \text{same terms} \right] + (72x^4 + 3x^3 - 6x^2 + 2x + 1)
- \left[ \text{same terms} \right] - (72x^4 + 4x^3 - 11x^2 - 6x + 13)
$$

This results in:

$$
f(x) - g(x) = 0 + (3x^3 - 4x^3) + (-6x^2 + 11x^2) + (2x + 6x) + (1 - 13)
$$

Simplifying:

$$
f(x) - g(x) = -x^3 + 5x^2 + 8x - 12
$$

Thus, the equation for intersection becomes:

$$
-x^3 + 5x^2 + 8x - 12 = 0
$$

Or, multiplying both sides by $-1$, we get:

$$
x^3 - 5x^2 - 8x + 12 = 0
$$

---

### Step 1: Solve the Cubic Equation

We attempt to factor this cubic. By **Rational Root Theorem**, the rational roots of the equation $ x^3 - 5x^2 - 8x + 12 = 0 $ are among the divisors of the constant term, $ \pm1, \pm2, \pm3, \pm4, \pm6, \pm12 $.

Testing $ x = 1 $:

$$
1^3 - 5(1)^2 - 8(1) + 12 = 1 - 5 - 8 + 12 = 0
\Rightarrow \text{a root!}
$$

So, we can factor the cubic as $ (x - 1)(x^2 - 4x - 12) $. Using the quadratic formula on $ x^2 - 4x - 12 $:

$$
x = \frac{4 \pm \sqrt{16 + 48}}{2} = \frac{4 \pm 8}{2}
\Rightarrow x = 6 \text{ or } x = -2
$$

Thus, the roots are:

$$
x = -2, \quad x = 1, \quad x = 6
$$

---

### Step 2: Determine Largest $ x $-Value

Since we are asked for the **largest** $ x $-value at which the graphs intersect, the only candidate is:

$$
\boxed{6}
$$

---

### Step 3: Verify the Validity of the Roots in Context

The presence of the **floor function** in the original expressions can make one question whether all cubic roots are valid. However, **any real $ x $** will have $ \lfloor x \rfloor $ equal to an integer $ n $, and the floor-dependent expressions (exponential, binomial coefficient, totient function) will only depend on that integer. Importantly, since all such terms are **identical** in both $ f(x) $ and $ g(x) $, they **will always cancel out** when taking $ f(x) - g(x) $.

This ensures that the **difference** between $ f(x) $ and $ g(x) $ is **always** equal to the polynomial $ -x^3 + 5x^2 + 8x - 12 $, regardless of the value of $ \lfloor x \rfloor $. Hence, **all real solutions** of the cubic equation are valid intersection points of the graphs of $ f(x) $ and $ g(x) $.

---

### Conclusion

The three roots of the cubic equation — $ x = -2, 1, 6 $ — are all valid intersection points between the two functions. Of these, the **largest** is:

$$
\boxed{6}
$$

Solution: The messy parts of the functions are irrelevant. All that matters for intersection is whether $f(x)-g(x)=0$. As $g(x)-f(x)=x^3-5x^2-8x+12=(x-6)(x+2)(x-1)$, the largest value of $x$ at which the graphs intersect is $x=\boxed{6}$.