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

1 Answer
Jul 3, 2016

Answer:

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.