Use the Inverse Function Theorem (Theorem 2.3 in the lecture notes) to show that f and g are inverses
of each other.
f (x) = x^3 + 1 g(x) = cubic root of
√(x − 1)

To show that \( f(x) = x^3 + 1 \) and \( g(x) = \sqrt[3]{x - 1} \) are inverses of each other using the **Inverse Function Theorem**, we'll follow these steps:

### 1. Check the compositions: \( f(g(x)) \) and \( g(f(x)) \)

For two functions to be inverses, their compositions must satisfy:

\[
f(g(x)) = x \quad \text{and} \quad g(f(x)) = x.
\]

Let's compute each composition:

#### \( f(g(x)) \)
We are given \( g(x) = \sqrt[3]{x - 1} \). Now, let's substitute \( g(x) \) into \( f(x) = x^3 + 1 \):

\[
f(g(x)) = f\left(\sqrt[3]{x - 1}\right) = \left( \sqrt[3]{x - 1} \right)^3 + 1.
\]

Using the property of cube roots, \( \left( \sqrt[3]{x - 1} \right)^3 = x - 1 \). Thus:

\[
f(g(x)) = (x - 1) + 1 = x.
\]

#### \( g(f(x)) \)
Now, substitute \( f(x) = x^3 + 1 \) into \( g(x) = \sqrt[3]{x - 1} \):

\[
g(f(x)) = g(x^3 + 1) = \sqrt[3]{(x^3 + 1) - 1} = \sqrt[3]{x^3}.
\]

Since \( \sqrt[3]{x^3} = x \), we have:

\[
g(f(x)) = x.
\]

### 2. Apply the **Inverse Function Theorem**

The **Inverse Function Theorem** states that if \( f \) is continuously differentiable and its derivative \( f'(x) \) is nonzero at a point, then \( f \) has a local inverse near that point. Moreover, the derivative of the inverse function \( g(x) \) at a point \( x \) can be expressed as:

\[
g'(x) = \frac{1}{f'(g(x))}.
\]

#### Check if \( f'(x) \) is nonzero

First, compute \( f'(x) \), the derivative of \( f(x) = x^3 + 1 \):

\[
f'(x) = 3x^2.
\]

Since \( 3x^2 \) is nonzero for all \( x \neq 0 \), \( f \) is invertible for \( x \neq 0 \) by the Inverse Function Theorem.

#### Verify using the Inverse Function Theorem

If \( f \) and \( g \) are inverses, we expect \( g'(x) = \frac{1}{f'(g(x))} \).

- First, find \( g'(x) \). Since \( g(x) = \sqrt[3]{x - 1} \), use the chain rule:

\[
g'(x) = \frac{1}{3} \cdot (x - 1)^{-\frac{2}{3}}.
\]

- Now, check \( \frac{1}{f'(g(x))} \):

Since \( f'(x) = 3x^2 \), substitute \( g(x) = \sqrt[3]{x - 1} \):

\[
f'(g(x)) = 3 \left( \sqrt[3]{x - 1} \right)^2 = 3 (x - 1)^{\frac{2}{3}}.
\]

Thus:

\[
\frac{1}{f'(g(x))} = \frac{1}{3 (x - 1)^{\frac{2}{3}}}.
\]

This matches \( g'(x) = \frac{1}{3} (x - 1)^{-\frac{2}{3}} \), confirming that \( f \) and \( g \) are indeed inverses of each other by the Inverse Function Theorem.

### Conclusion

We have shown that \( f(x) = x^3 + 1 \) and \( g(x) = \sqrt[3]{x - 1} \) are inverses of each other both by checking their compositions and verifying the conditions of the Inverse Function Theorem.

---

Would you like further details or have any specific questions on this?

Here are some related questions:

1. How does the Inverse Function Theorem guarantee the existence of an inverse?
2. What happens if \( f'(x) = 0 \) at a certain point?
3. Can we apply the Inverse Function Theorem to functions of multiple variables?
4. What role does continuity play in the Inverse Function Theorem?
5. How would the analysis change if \( f(x) \) had a more complex form, like \( f(x) = x^4 + x \)?

**Tip**: When using the Inverse Function Theorem, always ensure that the derivative of the function is nonzero at the points of interest.

Use the Inverse Function Theorem (Theorem 2.3 in the lecture notes) to show that f and g are inverses of each other. f(x) = x^3 + 1, g(x) = cubic root of √(x − 1)

Explore the Inverse Function Theorem and learn how to prove that f(x) = x^3 + 1 and g(x) = cubic root of √(x - 1) are inverses of each other. This detailed solution includes derivative calculations and function compositions. Ideal for undergraduate students in Calculus II or III.

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