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Chapter 0: Problem 3

Show that \(f\) and \(g\) are inverses of each other by verifying that \(f[g(x)]=x\) and \(g[f(x)]=x\). $$ f(x)=2 x+3 ; \quad g(x)=\frac{x-3}{2} $$

Short Answer

Expert verified
We have \(f(x) = 2x + 3\) and \(g(x) = \frac{x-3}{2}\). Plugging \(g(x)\) into \(f(x)\) gives \(f[g(x)] = 2(\frac{x-3}{2}) + 3 = x\). Plugging \(f(x)\) into \(g(x)\) gives \(g[f(x)] = \frac{2x + 3 - 3}{2} = x\). Since both \(f[g(x)] = x\) and \(g[f(x)] = x\), we can conclude that \(f(x)\) and \(g(x)\) are inverses of one another.

Step by step solution

01

Find \(f[g(x)]\)

To find \(f[g(x)]\), we need to plug in \(g(x)\) into the function \(f(x)\). We have \(f(x) = 2x+3\), and \(g(x) = \dfrac{x-3}{2}\). So we can write \(f[g(x)]\) as $$ f\left[\frac{x-3}{2}\right]=2 \left(\frac{x-3}{2}\right)+3 $$
02

Simplify \(f[g(x)]\)

Now we need to simplify the expression by applying arithmetic operations: $$ f\left[\frac{x-3}{2}\right] = 2 \cdot \frac{x-3}{2} + 3 \\ f\left[\frac{x-3}{2}\right] = (x-3) + 3 $$ After simplification, we obtain: $$ f\left[\frac{x-3}{2}\right] = x $$
03

Find \(g[f(x)]\)

To find \(g[f(x)]\), we need to plug in \(f(x)\) into the function \(g(x)\). We have \(g(x) = \dfrac{x - 3}{2}\), and \(f(x) = 2x + 3\). So we can write \(g[f(x)]\) as $$ g[2x + 3] = \frac{2x + 3 - 3}{2} $$
04

Simplify \(g[f(x)]\)

Now we need to simplify the expression by applying arithmetic operations: $$ g[2x + 3] = \frac{2x}{2} $$ After simplification, we obtain: $$ g[2x + 3] = x $$ Since we have shown that both \(f[g(x)] = x\) and \(g[f(x)] = x\), we can conclude that \(f(x)\) and \(g(x)\) are indeed inverses of one another.

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Most popular questions from this chapter

Suppose that \(f\) is a one-to-one function such that \(f(3)=7\) Find \(f\left[f^{-1}(7)\right]\).

You are given the graph of a function \(f .\) Determine whether \(f\) is one-to- one.

a. If \(f(x)=x-1\) and \(h(x)=2 x+3\), find a function \(g\) such that \(h=g \circ f\). b. If \(g(x)=3 x+4\) and \(h(x)=4 x-8\), find a function \(f\) such that \(h=g \circ f\).

Find the zero(s) of the function f to five decimal places. $$ f(x)=\sin 2 x-x^{2}+1 $$

Plot the graph of the function \(f\) in (a) the standard viewing window and (b) the indicated window. $$ f(x)=x^{3}-20 x^{2}+8 x-10 ; \quad[-20,20] \times[-1200,100] $$
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