Question

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

Answer 1

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`