# What are the extrema of f(x)=-8x^2+x on [-4,8]?

Feb 9, 2016

Absolute minimum of $- 512$ at $x = 8$ and an absolute maximum of $\frac{1}{32}$ at $x = \frac{1}{16}$

#### Explanation:

When finding the extrema on an interval, there are two locations they could be: at a critical value, or at one of the endpoints of the interval.

To find the critical values, find the function's derivative and set it equal to $0$. Since $f \left(x\right) = - 8 {x}^{2} + x$, through the power rule we know that $f ' \left(x\right) = - 16 x + 1$. Setting this equal to $0$ leaves us with one critical value at $x = \frac{1}{16}$.

Thus, our locations for potential maxima and minima are at $x = - 4$, $x = \frac{1}{16}$, and $x = 8$. Find each of their function values:

$f \left(- 4\right) = - 8 {\left(- 4\right)}^{2} - 4 = \underline{- 132}$

$f \left(\frac{1}{16}\right) = - 8 {\left(\frac{1}{16}\right)}^{2} + \frac{1}{16} = - \frac{1}{32} + \frac{1}{16} = \underline{\frac{1}{32}}$

$f \left(8\right) = - 8 {\left(8\right)}^{2} + 8 = \underline{- 504}$

Since the highest value is $\frac{1}{32}$, this is the absolute maximum on the interval. Note that the maximum itself is $\frac{1}{32}$, but its location is at $x = \frac{1}{16}$. Likewise, the lowest value and absolute minimum is $- 512$, located at $x = 8$.

This is $f \left(x\right)$ graphed: you can see that its maxima and minima are indeed where we found.

graph{-8x^2+x [-4.1, 8.1, -550, 50]}