Problem 33 - Thomas Calculus

Question

Suppose that the differentiable function \(y=f(x)\) has an inverse and that the graph of \(f\) passes through the point \((2,4)\) and has a slope of 1\(/ 3\) there. Find the value of \(d f^{-1} / d x\) at \(x=4\) .

Step-by-Step Solution

Answer

The value of \( \frac{d f^{-1}}{d x} \) at \( x=4 \) is 3.

Step 1: Understanding the problem

We are given a differentiable function \( y=f(x) \) that has an inverse. The graph of \( f \) passes through the point \((2,4)\) and has a slope of \( \frac{1}{3} \) there. We need to find the derivative of the inverse function, \( \frac{d}{dx}f^{-1}\), at \( x=4 \).

Step 2: Applying the inverse function derivative formula

We use the formula for the derivative of the inverse function: \( \left(f^{-1}\right)'(x) = \frac{1}{f'(f^{-1}(x))} \). This formula tells us how to find the derivative of \( f^{-1} \) at a point.

Step 3: Identifying needed values

From the problem, \( f(2) = 4 \), which means \( f^{-1}(4) = 2 \). The slope of \( f(x) \) at \( x=2 \) is \( f'(2) = \frac{1}{3} \).

Step 4: Calculating the derivative at the point

Using the formula, substitute the known values: \( \left(f^{-1}\right)'(4) = \frac{1}{f'(f^{-1}(4))} = \frac{1}{f'(2)}. \) Given \( f'(2) = \frac{1}{3} \), we have \( \left(f^{-1}\right)'(4) = \frac{1}{\frac{1}{3}} = 3 \).

Key Concepts

Differentiable Function

A differentiable function is a function that has a derivative at every point in its domain. This means the function must be smooth, without any sharp corners or breaks.
For example, if we have a function like a curve or line that we can draw without lifting our pen, it's likely differentiable.
This smoothness ensures that, at every point, we can determine the rate of change or slope of the function, which is essentially what differentiation provides.

• This is crucial because it allows us to use calculus tools like derivatives effectively.
• If a function is differentiable, it is also continuous. However, the reverse isn't always true: Continuous functions may not always be differentiable if there are sharp turns.

In the problem, the given function is differentiable, meaning we can find its derivative anywhere in its domain, which helps us solve for the derivative of its inverse.

Inverse Function Derivative Formula

The inverse function derivative formula is a fundamental tool in calculus that allows us to find the derivative of the inverse of a function.
It is given by: \[ \left(f^{-1}\right)'(x) = \frac{1}{f'(f^{-1}(x))} \] This formula is essential when dealing with inverse functions, enabling us to compute their derivatives if we know the derivative of the original function.

• The concept of an inverse function is straightforward: it "undoes" what the original function does. For instance, if \( f(x) \) turns \( x \) into \( y \), then \( f^{-1}(x) \) takes \( y \) back to \( x \).
• To find \( \left(f^{-1}\right)'(x) \), the formula requires you to compute the derivative of the inverse function at a particular point, using the reciprocal of the derivative of the original function at the corresponding point.

In our specific problem, we applied this formula to find the derivative of the inverse at \(x=4\), which helps us conclude that the slope of the inverse function there is 3.

Derivative Calculation

Calculating derivatives is a fundamental part of calculus, allowing us to determine how a function changes at any given point.
The derivative of a function \( f(x) \), denoted \( f'(x) \), describes the rate of change or the slope of the function.
In the given exercise, we have a function \( f \) with a known slope of \( \frac{1}{3} \) at the point \( (2, 4) \). This means at that point, the function is rising slowly.

• To compute the derivative of the inverse function at a given \( x \), we use the reciprocal of \( f'(x) \).
• So, when \( f'(2) = \frac{1}{3} \), the derivative of its inverse at the corresponding point is \( \frac{1}{\frac{1}{3}} = 3 \).

This computation tells us that the inverse function's graph is steeper at \( x=4 \), a useful insight when we explore the behaviors and relationships of functions and their inverses.