To solve for $g'(1)$ given the functions $f(x) = x^3 + \frac{e^x}{2}$ and $g[f(x)] = x$, we proceed as follows:

Step 1: Differentiate $g[f(x)] = x$ with respect to $x$

Using the chain rule: [ \frac{d}{dx}[g(f(x))] = \frac{d}{dx}[x] ] $g'(f(x)) \cdot f'(x) = 1$

Step 2: Solve for $g'(f(x))$

$g'(f(x)) = \frac{1}{f'(x)}$

Step 3: Evaluate $f'(x)$

The derivative of $f(x) = x^3 + \frac{e^x}{2}$ is: $f'(x) = 3x^2 + \frac{e^x}{2}$

Step 4: Identify $f(1)$

Substitute $x = 1$ into $f(x)$: $f(1) = 1^3 + \frac{e^1}{2} = 1 + \frac{e}{2} = 1 + \frac{e}{2}$

Step 5: Solve for $g'(1)$

Since $g[f(1)] = 1$, we know $f(1) = 1 + \frac{e}{2}$. Using the formula for $g'(f(x))$: $g'(1) = g'(f(1)) = \frac{1}{f'(1)}$

Substitute $x = 1$ into $f'(x)$: $f'(1) = 3(1^2) + \frac{e^1}{2} = 3 + \frac{e}{2}$

Thus: $g'(1) = \frac{1}{f'(1)} = \frac{1}{3 + \frac{e}{2}}$

Final Answer:

$g'(1) = \frac{2}{6 + e}$

Let me know if you'd like a detailed explanation of any step.

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Related Questions:

1. What is the general expression for $g'(x)$ in terms of $f(x)$?
2. How does the chain rule work for nested functions?
3. What happens if $f(x)$ changes to $x^3 - \frac{e^x}{2}$?
4. Can $g'(x)$ ever equal zero? Why or why not?
5. How would this problem change if $f'(x)$ had a critical point at $x = 1$?

Tip:

Always ensure that $f(x)$ is invertible when solving such problems, as $g[f(x)] = x$ implicitly assumes $f(x)$ has an inverse.