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Chapter 0: Problem 31

Find the inverse of \(f .\) Then use a graphing utility to plot the graphs of \(f\) and \(f^{-1}\) using the same viewing window. $$ f(x)=\frac{x}{x^{2}+1}, \quad-\frac{1}{2} \leq x \leq \frac{1}{2} $$

Short Answer

Expert verified
Our inverse function is given as \(f^{-1}(x) = \frac{1 + \sqrt{1 - 4x^2}}{2x}\). After plotting both \(f(x) = \frac{x}{x^2 + 1}\) and \(f^{-1}(x)\), you should notice that they are symmetric around the line \(y = x\).

Step by step solution

01

Replace f(x) with y

Write the given function in terms of y: \[y = \frac{x}{x^2 + 1}\]
02

Swap x and y

Switch the roles of x and y in the equation: \[x = \frac{y}{y^2 + 1}\]
03

Solve for y

To solve for y, we'll first multiply both sides by \((y^2 + 1)\): \[x(y^2 + 1) = y\] Now, distribute the x: \[xy^2 + x = y\] Move all terms to one side of the equation: \[xy^2 - y + x = 0\] This is a quadratic equation in terms of y. To find the inverse, we need to solve this equation for y using the quadratic formula, which is: \[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] In our equation, we have \(a = x\), \(b = -1\), and \(c = x\). Plug these values into the quadratic formula: \[y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(x)(x)}}{2(x)}\]
04

Find the inverse function, f^{-1}(x)

Simplify the expression: \[y = \frac{1 \pm \sqrt{1 - 4x^2}}{2x}\] To get a valid function as an inverse, we have to select one of the possible signs between plus and minus. Since we're in the interval \(-\frac{1}{2} \leq x \leq \frac{1}{2}\), we'll choose the positive sign so that our inverse function has a continuous domain. Our inverse function is given as: \[f^{-1}(x) = \frac{1 + \sqrt{1 - 4x^2}}{2x}\]
05

Plot the graphs

Now, we should plot both functions, \(f(x)\) and \(f^{-1}(x)\), in the same viewing window of the graphing utility: 1. \(f(x) = \frac{x}{x^2 + 1}\): This is the original function given. 2. \(f^{-1}(x) = \frac{1 + \sqrt{1 - 4x^2}}{2x}\): This is the inverse function we found. Once you have plotted both graphs, you should notice that they are symmetric around the line \(y = x\).

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Most popular questions from this chapter

Prove that a function has an inverse if and only if it is oneto-one.

Sketch the graph of the first function by plotting points if necessary. Then use transformation(s) to obtain the graph of the second function. \(y=\sin x, \quad y=2 \sin \frac{x}{2}\)

Find the exact value of the given expression. $$ \cos ^{-1} 0 $$

Find the exact value of the given expression. $$ \sin ^{-1} 0 $$

Find \(f^{-1}(a)\) for the function \(f\) and the real number \(a\). $$ f(x)=\frac{3}{\pi} x+\sin x ; \quad-\frac{\pi}{2}
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