# Question #b6e1b

Jan 24, 2018

Yes. See explanation.

#### Explanation:

If $f \left(x\right) = \frac{1}{2} x + 1$, then we know that $f \left(x\right)$ is increasing with a constant slope of $\frac{1}{2}$. That means that $f \left(x\right)$ is a one-to-one function, so it has an inverse that is a function.

You could also just find the inverse in this case. If $g \left(x\right) = {f}^{-} 1 \left(x\right)$:

$f \left(g \left(x\right)\right) = \frac{1}{2} g \left(x\right) + 1 = x$

Solving for $g \left(x\right)$:

$\frac{1}{2} g \left(x\right) + 1 = x$

subtract 1 from each side:

$\frac{1}{2} g \left(x\right) = x - 1$

multiply through by 2:
$g \left(x\right) = 2 x - 2$