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Chapter 1: Problem 23

Verify that \(f\) and \(g\) are inverse functions (a) algebraically and (b) graphically. $$f(x)=7 x+1, \quad g(x)=\frac{x-1}{7}$$

Short Answer

Expert verified
Algebraically, the results of both \(f(g(x))\) and \(g(f(x))\) are \(x\), proving that \(f\) and \(g\) are inverses. Graphically, these functions are mirror images of each other when reflected over the line \(y = x\), further verifying their inverse relationship.

Step by step solution

01

Composition of Functions

In the first step, start by calculating the composition of \(f\) and \(g\). We need to show both \(f(g(x))\) and \(g(f(x))\) return \(x\). For \(f(g(x))\), substitute \(g(x)\) into \(f(x)\) to get \(f(g(x)) = f((x-1)/7) = 7((x-1)/7) + 1 = x \).For \(g(f(x))\), substitute \(f(x)\) into \(g(x)\) to get \(g(f(x)) = g(7x + 1) = (7x + 1 - 1) / 7 = x\). Since both \(f(g(x))\) and \(g(f(x))\) result in \(x\), it confirms that \(f\) and \(g\) are inverses of each other algebraically.
02

Graphical Representation

For the graphical interpretation, plot both \(f(x)=7x+1\) and \(g(x)=(x-1)/7\) on the same graph. Then, draw the line \(y = x\). If \(f\) and \(g\) are inverses, they will reflect over the line \(y = x\). In this case, the line \(g(x) = (x-1)/7\) is a reflection of \(f(x) = 7x + 1\) over the line \(y = x\), confirming that \(f\) and \(g\) are indeed inverses graphically.

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