Question

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

Answer 1

Absolute minimum of `-512` at `x=8` and an absolute maximum of `1/32` at `x=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(x)=-8x^2+x`, through the power rule we know that `f'(x)=-16x+1`. Setting this equal to `0` leaves us with one critical value at `x=1/16`.

Thus, our locations for potential maxima and minima are at `x=-4`, `x=1/16`, and `x=8`. Find each of their function values:

`f(-4)=-8(-4)^2-4=ul(-132)`

`f(1/16)=-8(1/16)^2+1/16=-1/32+1/16=ul(1/32)`

`f(8)=-8(8)^2+8=ul(-504)`

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

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

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