To solve the problem, we will use the relationship between a function and its inverse:

$f(f^{-1}(x)) = x \quad \text{and hence} \quad (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}.$

Here, we are tasked to find $(f^{-1})'(0)$. The function $f(x)$ is defined as:

$f(x) = \int_{\pi}^x \frac{2 + e^{\sin t}}{3} \, dt.$

Step 1: Evaluate $f'(x)$

Using the Fundamental Theorem of Calculus, the derivative of $f(x)$ is:

$f'(x) = \frac{2 + e^{\sin x}}{3}.$

Step 2: Find $f(c) = 0$

Since $f(x)$ has an inverse, we need to determine $c$ such that $f(c) = 0$. From the definition of $f(x)$:

$f(c) = \int_{\pi}^c \frac{2 + e^{\sin t}}{3} \, dt = 0.$

This implies $c = \pi$, because the integral from $\pi$ to itself is $0$.

Thus, $f^{-1}(0) = \pi$.

Step 3: Compute $(f^{-1})'(0)$

From the inverse function derivative formula:

$(f^{-1})'(0) = \frac{1}{f'(f^{-1}(0))}.$

Since $f^{-1}(0) = \pi$, this becomes:

$(f^{-1})'(0) = \frac{1}{f'(\pi)}.$

Using $f'(x) = \frac{2 + e^{\sin x}}{3}$, substitute $x = \pi$:

$f'(\pi) = \frac{2 + e^{\sin \pi}}{3}.$

Since $\sin \pi = 0$, we have:

$f'(\pi) = \frac{2 + e^0}{3} = \frac{2 + 1}{3} = \frac{3}{3} = 1.$

Thus:

$(f^{-1})'(0) = \frac{1}{f'(\pi)} = \frac{1}{1} = 1.$

Final Answer:

The derivative $(f^{-1})'(0)$ is:

(a) 1.

Would you like a detailed breakdown of any step or related questions? 😊

Here are 5 follow-up questions for deeper understanding:

1. How do you apply the Fundamental Theorem of Calculus to evaluate $f'(x)$?
2. Why does the inverse function formula depend on $f'(f^{-1}(x))$?
3. Can $f(x)$ have an inverse if its derivative $f'(x)$ is not continuous?
4. How do we interpret the result $(f^{-1})'(x) = 1/f'(f^{-1}(x))$ geometrically?
5. How would this problem change if the bounds of the integral were $0$ to $x$ instead of $\pi$ to $x$?

Tip: Always verify the domain and monotonicity of a function when determining if it has an inverse.