To solve this problem, let's find $f''(x)$ at $x = 1$, given:

$f(x) = 2e - \cos(x - 1) + \frac{1}{e^x} - \sqrt[3]{x}.$

Step 1: Compute the first derivative, $f'(x)$

The function is a sum of four terms, so we differentiate each term separately:

1. $2e$: This is a constant, so its derivative is $0$.
2. $-\cos(x - 1)$: The derivative is $\sin(x - 1)$.
3. $\frac{1}{e^x}$: Rewrite as $e^{-x}$, and the derivative is $-e^{-x}$.
4. $-\sqrt[3]{x}$: Rewrite as $-x^{1/3}$, and the derivative is $-\frac{1}{3}x^{-2/3}$.

Thus: $f'(x) = \sin(x - 1) - e^{-x} - \frac{1}{3}x^{-2/3}.$

Step 2: Compute the second derivative, $f''(x)$

Now, differentiate $f'(x)$ term by term:

1. $\sin(x - 1)$: The derivative is $\cos(x - 1)$.
2. $-e^{-x}$: The derivative is $e^{-x}$.
3. $-\frac{1}{3}x^{-2/3}$: The derivative is $\frac{2}{9}x^{-5/3}$.

Thus: $f''(x) = \cos(x - 1) + e^{-x} + \frac{2}{9}x^{-5/3}.$

Step 3: Evaluate $f''(x)$ at $x = 1$

Substitute $x = 1$ into $f''(x)$:

1. $\cos(1 - 1) = \cos(0) = 1$,
2. $e^{-1} = \frac{1}{e}$,
3. $\frac{2}{9} \cdot 1^{-5/3} = \frac{2}{9}$.

Thus: $f''(1) = 1 + \frac{1}{e} + \frac{2}{9}.$

Final Answer:

$f''(1) = 1 + \frac{1}{e} + \frac{2}{9}.$

Would you like further simplification or explanations? Here are some related questions to deepen your understanding:

1. How do we find higher-order derivatives for composite functions like $\cos(x - 1)$?
2. What role does $e$ play in differentiation?
3. Can the cube root term $\sqrt[3]{x}$ create undefined derivatives for $x \leq 0$?
4. Why is $e^x$ unique among exponential functions in calculus?
5. How do we handle fractional powers when computing derivatives?

Tip: Always simplify the function before differentiating to minimize errors!