What are the local extrema, if any, of f (x) = x^3 - 6x^2 - 15x + 11 ?

1 Answer
Nov 28, 2015

Maxima=19 at x=-1
Minimum=-89 atx=5

Explanation:

f(x) = x^3-6x^2-15x+11

To find the local extrema first find the critical point

f'(x) = 3x^2-12x-15

Set f'(x)=0

3x^2-12x-15=0

3(x^2-4x-5)=0

3(x-5)(x+1)=0

x=5 or x=-1 are critical points. We need to do the second derivative test

f^('')(x)=6x-12

f^('')(5)=18 >0 , so f attains its minimum at x=5 and the minimum value is f(5)=-89

f^('')(-1) = -18 < 0 , so f attains its maximum at x=-1 and the maximum value is f(-1)=19