How do you find the critical points for #f(x) = x^3 - 15x^2 + 4# and the local max and min?

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Answer 1 · 2016-12-21T15:15:01

The critical points are: #x_1=0# and #x_2=10#.
#x_1# is a maximum, while #x_2# is a minimum.

To identify the critical points, we have to solve the equation:

#f'(x)=0#

that is:

#3x^2-30x=0#

#3x(x-10) =0#

so that the critical points for #f(x)# are #x_1=0# and #x_2=10#.
We can now calculate the second derivative:

#f''(x) = 6x-30#

and note that around #x_1# #f''(x)# is negative, therefore #x_1# is a maximum, while around #x_2# it is positive, therefore #x_2# is a minimum.

graph{x^3-15x^2+4 [-80, 80, -640, 640]}