What are the absolute extrema of f(x) =x^4 − 8x^2 − 12 in[-3,-1]?

1 Answer
May 30, 2016

-3 (occurring at x=-3) and -28 (occurring at x=-2)

Explanation:

Absolute extrema of a closed interval occur at the endpoints of the interval or at f'(x)=0.

That means we'll have to set the derivative equal to 0 and see what x-values that gets us, and we'll have to use x=-3 and x=-1 (because these are the endpoints).

So, starting with taking the derivative:
f(x)=x^4-8x^2-12
f'(x)=4x^3-16x

Setting it equal to 0 and solving:
0=4x^3-16x
0=x^3-4x
0=x(x^2-4)
x=0 and x^2-4=0
Thus the solutions are 0,2, and -2.

We immediately get rid of 0 and 2 because they are not on the interval [-3,-1], leaving only x=-3,-2, and -1 as the possible places where extrema can occur.

Finally, we evaluate these one by one to see what the absolute min and max are:
f(-3)=-3
f(-2)=-28
f(-1)=-19

Therefore -3 is the absolute maximum and -28 is the absolute minimum on the interval [-3,-1].