Problem 9 - University Calculus: Early Transcendentals

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 function \(y = f(x)\) that is differentiable and has an inverse. The function passes through the point (2,4), which means \(f(2) = 4\), and the slope at this point is \(\frac{1}{3}\). We need to find the derivative of its inverse \(f^{-1}(x)\) at \(x = 4\).

Step 2: Applying Derivative of Inverse Function

The derivative of an inverse function at a point is given by the formula:\[ \left( f^{-1} \right)'(b) = \frac{1}{f'(a)} \]where \(f(a) = b\). In this problem, \(f(2) = 4\), so \(a = 2\) and \(b = 4\). The derivative of \(f\) at \(x = 2\) is \(\frac{1}{3}\).

Step 3: Calculate Inverse Derivative

Using the derivative formula for inverse functions, substitute \(a = 2\) where the derivative \(f'(2) = \frac{1}{3}\) and \(b = 4\):\[ \left( f^{-1} \right)'(4) = \frac{1}{f'(2)} = \frac{1}{\frac{1}{3}} = 3 \].

Key Concepts

Differentiable Function

To begin with, it’s essential to understand what a differentiable function is. Differentiability means that a function has a derivative at each point in its domain.
This implies the function is smooth and does not have any sharp corners or cusps. In mathematical terms, if a function is differentiable at a point, its derivative exists at that point. If a function is differentiable everywhere in its domain, it means you can find its derivative for any value within that domain. A common example of a differentiable function is a polynomial, which is smooth and continuous.
In our exercise, the function given is differentiable and has an inverse, indicating its smoothness and continuity as we pass through the specific point given in the problem.

Derivative of Inverse Function

The derivative of an inverse function can be slightly tricky, but once you know the rule, it's straightforward. The main principle relies on the relationship between the derivatives of a function and its inverse.
If you have a function, say, \( y = f(x) \), with its inverse \( x = f^{-1}(y) \), then at any point \( (a, b) \) where \( f(a) = b \) and \( f^{-1}(b) = a \), the derivative of the inverse of \( f \) at \( b \) is:

• \( \left( f^{-1} \right)'(b) = \frac{1}{f'(a)} \)

This rule is significant because it allows us to determine how a small change in \( y \) affects \( x \). The inverse relation of derivatives provides insights into how sensitive the inverse function is to changes at specific points.

Inverse Derivative Calculation

Calculating the derivative of an inverse function becomes an easy process with the right formula. In our exercise, we have the function \( f(x) \), which passes through the point (2,4) with a slope of \( \frac{1}{3} \).First, identify the values for \( a \) and \( b \) using the information given:

• \( f(2) = 4 \) ➔ \( a = 2 \) and \( b = 4 \)
• The slope or derivative at \( x = 2 \) is \( \frac{1}{3} \)

Now apply these into the formula for the derivative of an inverse:

• \( \left( f^{-1} \right)'(4) = \frac{1}{f'(2)} = \frac{1}{\frac{1}{3}} = 3 \)

This calculation shows that at \( x = 4 \), the inverse function's slope is 3. This means that for a small change in \( x \) near 4, \( f^{-1}(x) \) changes three times as much.