To solve these problems, we need to use the fact that the derivative of the inverse function at a given point $a$ is:

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

We'll proceed with each question individually.

Problem 29

Given: $f(x) = x^3 + 2$ and $a = -1$.

1. Verify that $f$ has an inverse:
   Since $f(x) = x^3 + 2$ is a strictly increasing function (its derivative $f'(x) = 3x^2$ is positive for all $x \neq 0$), $f$ is one-to-one and hence has an inverse.

2. Find $f^{-1}(a)$: We need to solve for $x$ such that $f(x) = -1$: $x^3 + 2 = -1$ $x^3 = -3$ $x = -\sqrt[3]{3}$ So, $f^{-1}(-1) = -\sqrt[3]{3}$.

3. Find $f'(x)$: $f'(x) = 3x^2$

4. Evaluate $f'(f^{-1}(a))$: Substitute $x = -\sqrt[3]{3}$ into $f'(x)$: [ f'(f^{-1}(-1)) = f'(-\sqrt[3]{3}) = 3(-\sqrt[3]{3})^2 = 3 \cdot \sqrt[3]{9} ]

5. Calculate $(f^{-1})'(-1)$: $(f^{-1})'(-1) = \frac{1}{3 \cdot \sqrt[3]{9}}$

So, the answer for problem 29 is: $(f^{-1})'(-1) = \frac{1}{3 \cdot \sqrt[3]{9}}$

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Problem 31

Given: $f(x) = \tan x, \quad -\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$ and $a = \frac{\sqrt{3}}{3}$.

1. Verify that $f$ has an inverse:
   Within the interval $-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$, $f(x) = \tan x$ is strictly increasing and continuous, thus it is one-to-one and has an inverse.

2. Find $f^{-1}(a)$: We need to solve for $x$ such that $f(x) = \frac{\sqrt{3}}{3}$: $\tan x = \frac{\sqrt{3}}{3}$ The solution within the given interval is: $x = \frac{\pi}{6}$ Thus, $f^{-1}\left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6}$.

3. Find $f'(x)$: $f'(x) = \sec^2 x$

4. Evaluate $f'(f^{-1}(a))$: Substitute $x = \frac{\pi}{6}$ into $f'(x)$: $f'\left(f^{-1}\left(\frac{\sqrt{3}}{3}\right)\right) = f'\left(\frac{\pi}{6}\right) = \sec^2\left(\frac{\pi}{6}\right)$ Since $\sec\left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}}$, $\sec^2\left(\frac{\pi}{6}\right) = \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3}$

5. Calculate $\left(f^{-1}\right)'\left(\frac{\sqrt{3}}{3}\right)$: $\left(f^{-1}\right)'\left(\frac{\sqrt{3}}{3}\right) = \frac{1}{\frac{4}{3}} = \frac{3}{4}$

So, the answer for problem 31 is: $\left(f^{-1}\right)'\left(\frac{\sqrt{3}}{3}\right) = \frac{3}{4}$

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Would you like a further breakdown of these steps, or do you have any other questions?

Related Questions

1. How can we generalize the formula for the derivative of an inverse function?
2. Why is the domain restriction necessary for $f(x) = \tan x$ to have an inverse?
3. Can you explain why a function being strictly increasing implies it has an inverse?
4. How would the derivative of the inverse change if the function were decreasing instead of increasing?
5. What would be the steps to find the inverse derivative of a composite function?

Tip

When verifying if a function has an inverse, checking if it is strictly monotonic (either increasing or decreasing) over the entire domain can simplify the process.