Question

Find the absolute minimum and maximum values of the given interval. f(x)=(x^2-1)^3, [-1,5]

Answer 1

First find the derivative. When you found this try f'(x) = 0.

Answer 2

6x(x^2-1)^2

Answer 3

Now what??

Answer 4

To evaluate the equation 6x(x^2-1)^2 you have got to split it in 2 parts. Namely 6x and (x^2-1)^2. 6x=0 gives x=0 and (x^2-1)^2=0 gives x=1. This means that there are two points on this line where the slope of f(x) is 0 (and hence there is an minimum or maximum). Just filling in there values in the original function gives f(0) = -1 and f(1) = 0. So now you know where the lowest point on this interval is.

Answer 5

Did this answer your question?

Answer 6

It does, Thank you. The absolute minimum is -1. The max f(5)= 13,824-which seems rather large? Is this correct?

Answer 7

\(f(5)\) should be rather large because it is \(24^3\)

Answer 8

Yes you're right. Meaning this is a maximum on this interval. I only helped KJD finding the absolut minimum. This is my bad btw.

Answer 9

Min=-1; Max=13,824. It was rather large but correct. Thank you for your guidance!