Question

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

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

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 $$ f\left[\frac{x-3}{2}\right]=2 \left(\frac{x-3}{2}\right)+3 $$

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: $$ f\left[\frac{x-3}{2}\right] = x $$

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 $$ g[2x + 3] = \frac{2x + 3 - 3}{2} $$

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.