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

Nov 28, 2015

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

#### Explanation:

$f \left(x\right) = {x}^{3} - 6 {x}^{2} - 15 x + 11$

To find the local extrema first find the critical point

$f ' \left(x\right) = 3 {x}^{2} - 12 x - 15$

Set $f ' \left(x\right) = 0$

$3 {x}^{2} - 12 x - 15$=0

$3 \left({x}^{2} - 4 x - 5\right)$=0

$3 \left(x - 5\right) \left(x + 1\right) = 0$

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

${f}^{' '} \left(x\right) = 6 x - 12$

${f}^{' '} \left(5\right) = 18 > 0$ , so $f$ attains its minimum at $x = 5$ and the minimum value is $f \left(5\right) = - 89$

${f}^{' '} \left(- 1\right) = - 18 < 0$ , so $f$ attains its maximum at $x = - 1$ and the maximum value is $f \left(- 1\right) = 19$