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 $x$.