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Chapter 3: Problem 9

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 \(x=4\).

Short Answer

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

Step by step solution

01

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 \(x = 4\).
02

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 \(\frac{1}{3}\).
03

Calculate Inverse Derivative

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiable Function
To begin with, it鈥檚 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:
  • \( f(2) = 4 \) 鉃� \( a = 2 \) 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.

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Most popular questions from this chapter

Measuring acceleration of gravity When the length \(L\) of a clock pendulum is held constant by controlling its temperature, the pendulum's period \(T\) depends on the acceleration of gravity \(g\). The period will therefore vary slightly as the clock is moved from place to place on the earth's surface, depending on the change in g. By keeping track of \(\Delta T\), we can estimate the variation in \(g\) from the equation \(T=2 \pi(L / g)^{1 / 2}\) that relates \(T, g,\) and \(L\)a. With \(L\) held constant and \(g\) as the independent variable, calculate \(d T\) and use it to answer parts (b) and (c). b. If \(g\) increases, will \(T\) increase or decrease? Will a pendulum clock speed up or slow down? Explain. c. A clock with a \(100-\mathrm{cm}\) pendulum is moved from a location where \(g=980 \mathrm{cm} / \mathrm{sec}^{2}\) to a new location. This increases the period by \(d T=0.001 \mathrm{sec} .\) Find \(d g\) and estimate the value of \(g\) at the new location.

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