https://file.aiquickdraw.com/mathbot/file/1729899887315n4q89f4f.jpg

To solve \( (f^{-1})'(5) \) where \( f(x) = x^3 + x + 3 \), we’ll use the formula for the derivative of the inverse function. If \( y = f(x) \), then

\[
(f^{-1})'(y) = \frac{1}{f'(x)}
\]

for \( y = f(x) \).

### Step 1: Set \( f(x) = 5 \)
We need to find the value of \( x \) such that \( f(x) = 5 \):
\[
x^3 + x + 3 = 5
\]
\[
x^3 + x - 2 = 0
\]
Testing \( x = 1 \):
\[
1^3 + 1 - 2 = 0
\]
So, \( x = 1 \) is a solution.

### Step 2: Find \( f'(x) \)
Now, we find the derivative of \( f(x) \):
\[
f(x) = x^3 + x + 3
\]
\[
f'(x) = 3x^2 + 1
\]

### Step 3: Evaluate \( f'(1) \)
Since \( x = 1 \), we evaluate \( f'(1) \):
\[
f'(1) = 3(1)^2 + 1 = 3 + 1 = 4
\]

### Step 4: Calculate \( (f^{-1})'(5) \)
Now, using the inverse derivative formula:
\[
(f^{-1})'(5) = \frac{1}{f'(1)} = \frac{1}{4}
\]

### Final Answer
\[
(f^{-1})'(5) = \frac{1}{4}
\]

Would you like more details or have any questions?

### Related Questions
1. How can we generalize the formula for the derivative of an inverse function?
2. What if \( f(x) = x^3 + x + c \)? How does \( c \) affect \( (f^{-1})'(y) \)?
3. What is the significance of finding \( f^{-1} \) derivatives in calculus?
4. How would we solve \( f(x) = y \) for other values of \( y \) with this function?
5. How do we know \( f(x) \) has an inverse for certain values of \( y \)?

### Tip
When dealing with inverse functions, remember that the derivative \( (f^{-1})'(y) = \frac{1}{f'(x)} \) only applies when \( f \) is one-to-one near the value of \( x \).

Find \((f^{-1})'(5)\) if \(f(x) = x^3 + x + 3\).

Learn how to find the derivative of the inverse function (f^-1)'(5) when f(x) = x^3 + x + 3. This solution covers the use of the inverse function theorem, including finding derivatives and setting up equations. Suitable for advanced calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation