Okay, let's try to figure out where the two functions f(x) and g(x) intersect. So, the problem is asking for the largest x-value where f(x) equals g(x). Let me start by understanding the functions given.

First, both f(x) and g(x) have a lot of similar terms. Let me write them out to compare:

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

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

Hmm, so the only differences between f(x) and g(x) are in the cubic, quadratic, linear, and constant terms. Let me subtract g(x) from f(x) to see where they might be equal:

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

Let me compute that term by term:

3x³ -4x³ = -x³

-6x² - (-11x²) = -6x² +11x² = 5x²

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

1 -13 = -12

So, f(x) - g(x) = -x³ +5x² +8x -12

We need to solve for x where this difference is zero:

-x³ +5x² +8x -12 =0

Multiply both sides by -1 to make the leading coefficient positive:

x³ -5x² -8x +12 =0

So, we need to find the roots of the cubic equation x³ -5x² -8x +12 =0.

Let me try to factor this cubic equation. 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= -4, -4-8=-12, -12+12=0. Yes! So x=1 is a root.

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

Using synthetic division with root 1:

Coefficients: 1 | -5 | -8 | 12

Bring down 1.

Multiply by 1: 1*1=1. Add to next coefficient: -5 +1= -4.

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

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

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

x² -4x -12. Let's find two numbers multiplying to -12 and adding to -4. Hmm, factors of -12:  -6 and 2. -6 +2= -4. Yes.

So, x² -4x -12 = (x -6)(x +2). Therefore, the cubic factors are (x -1)(x -6)(x +2).

Thus, the roots are x=1, x=6, x=-2.

But the problem asks for the largest x-value where f(x)=g(x). So among the roots x=1, x=6, x=-2, the largest is x=6.

But wait, we need to check if these roots are valid in the original functions. Because the functions involve the floor function |⌊x⌋|, which is an integer. Let me check for each root x=6, x=1, x=-2 whether the substitution into the floor function is valid.

First, x=6: floor(6)=6. Then |⌊6⌋|=6. Then 6! is 720. Then in the exponent of e, we have 3*(6)^2 -720= 3*36 -720=108 -720= -612. The exponent is negative, but e^{-612} is a very small positive number, which is fine. The binomial coefficient is C(22 +735235*6, 2356). Since 735235*6 is a huge number, adding 22, but the binomial coefficient C(n, k) is defined for n >=k. Since 735235*6 +22 is way larger than 2356, so that's okay. The Euler's totient function φ(6 +1)=φ(7)=6, since 7 is prime. The rest of the terms are polynomials in x, which are valid. So x=6 is valid.

x=1: floor(1)=1. |1|=1. 1! =1. Exponent in e is 3*(1)^2 -1=3 -1=2. e^2 is fine. The binomial coefficient is C(22 +735235*1,2356)=C(735257,2356), which is a huge number but computable. φ(1 +1)=φ(2)=1. The rest of the terms are polynomials. So x=1 is valid.

x=-2: floor(-2)= -2. |floor(-2)|=2. 2! =2. Exponent in e is 3*(-2)^2 -2= 12 -2=10. e^10 is okay. The binomial coefficient is C(22 +735235*2,2356)=C(1470492,2356), which is valid. φ(|floor(-2)| +1)=φ(2 +1)=φ(3)=2. The rest of the terms are polynomials in x, which are valid for x=-2. So x=-2 is valid.

Therefore, all three roots are valid. The largest x-value is 6. Therefore, the answer is 6.

But wait, let me check if there's any domain restrictions. For example, the floor function's argument is x, but since x is a real number, the floor function is defined for all real numbers. The exponentials, binomial coefficients, Euler's totient function, and polynomials are all defined for the given expressions. So, all three roots are valid.

Hence, the largest x is 6.

**Final Answer**
The largest $x$-value at which the graphs intersect is $\boxed{6}$.
Thus, the largest $x$-value where the two graphs intersect is $\boxed{6}$.
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