What are the local extrema of f(x) = x^3 - 3x^2 - 9x +1?

May 28, 2017

relative maximum: $\left(- 1 , 6\right)$
relative minimum: $\left(3 , - 26\right)$

Explanation:

Given: $f \left(x\right) = {x}^{3} - 3 {x}^{2} - 9 x + 1$

Find the critical numbers by finding the first derivative and setting it equal to zero:

$f ' \left(x\right) = 3 {x}^{2} - 6 x - 9 = 0$

Factor: $\left(3 x + 3\right) \left(x - 3\right) = 0$

Critical numbers: $x = - 1 , \text{ } x = 3$

Use the second derivative test to find out if these critical numbers are relative maximums or relative minimums:

$f ' ' \left(x\right) = 6 x - 6$

$f ' ' \left(- 1\right) = - 12 < 0 \implies \text{ relative max at } x = - 1$

$f ' ' \left(3\right) = 12 > 0 \implies \text{ relative min at } x = 3$

$f \left(- 1\right) = {\left(- 1\right)}^{3} - 3 {\left(- 1\right)}^{2} - 9 \left(- 1\right) + 1 = 6$

$f \left(3\right) = {3}^{3} - 3 {\left(3\right)}^{2} - 9 \left(3\right) + 1 = - 26$

relative maximum: $\left(- 1 , 6\right)$
relative minimum: $\left(3 , - 26\right)$