# Home, Home on the Range

The first week of school I gave this problem and never came back to it:

Find the range of $f(x)=2^{x^2-4x+1}$.

The answer is $\left[\frac{1}{8},\infty \right)$. Here’s why:

First consider the range of the quadratic in the exponent, $h(x)=x^2-4x+1$.  It’s a parabola that opens up with its vertex at $(2,-3)$. So the range of $h(x)$ is $[-3,\infty)$.

Now, consider the function $g(x)=2^x$.  If we let the domain of $g$ be the values coming from  the range of $h$, we have the mapping $\left[-3,\infty\right)\to\mathbb{R}$. That is, we’re considering the composition $f(x)=g(h(x))$. Since $g$ is monotonically increasing, for any $x_1, we know $g(x_1). So the range of $g$ in $\mathbb{R}$ is $\left[g(-3),\infty\right)$. So the range of $f$ is $\left[\frac{1}{8},\infty \right)$.

Do you feel “at home on the range”?

Here are a few more for you to try. In each case, find the range of the function. These aren’t meant to be any harder than the original problem, just different. Though watch out for the third one :-).

$y=3^{-x^2+4}$

$y=4^{\sqrt{x}}$

$y=\ln{(x^2-x+1)}$

$y=\sin{(x^2+2x+5)}$