How do you find the critical points for f(x) = (x^2-10x)^4 and the local max and min?

1 Answer
Maharshi
Jul 3, 2016

three critical points: x=0,5,10
Relative minimum at x=0, 10
Relative maximum at x=5

Explanation:

First, we need to find f'(x).
Using rules for taking derivative, we get following:
f'(x) = 4*(x^2-10x)^3*(2x-10)

Critical points occur when the slope is zero meaning f'(x)=0.

f'(x) = 4*(x^2-10x)^3*(2x-10)=0

(2x-10)=0 => x=5
(x^2-10x)=0 => x=0, 10

So, three critical points would be x=0,5,10.

To find relative maximum and minimum, we can do a sign test.

For x in(-∞, 0) => f'(x) is negative.
For x in(0, 5) => f'(x) is positive.
For x in(5, 10) => f'(x) is negative.
For x in(10, ∞) => f'(x) is positive.

At x=0, function has relative minimum because f'(x) changes signs from negative to positive.

At x=5, function has relative maximum because f'(x) changes signs from positive to negative.

At x=10, function has relative minimum because f'(x) changes signs from negative to positive.

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