The function $f(x)$ is defined only on domain $[-1,2]$, and is defined on this domain by the formula

Question

$$f(x) = 2x^2-8x+1.$$
What is the range of $f(x)$? Express your answer as an interval or as a union of intervals.

Answer 1

Completing the square, we obtain \begin{align*}
f(x)&=2\left(x-\frac{4}{2}\right)^2+1-2\left(\frac{4}{2}\right)^2 \\
&=2\left(x-2\right)^2+1-2(4) \\
&=2\left(x-2\right)^2-7.
\end{align*} Since $(x-2)^2\ge 0$, we have $2(x-2)^2\ge 0$. Therefore, $f(x) \ge 0 - 7 = -7$. Since $f(x)$ is a quadratic with a positive leading coefficient, we know $f(x)$ is minimized at its vertex. The vertex occurs when $x=2$, so when $x$ is in the domain $[-1,2]$, we know $f(x)$ is at most $2(0)^2 - 7 =-7$. Therefore, the range of $f(x)$ is $[-7, \infty)$.