# How do you find all local extrema/ relative maxima and minima for the function y = x^3 - 3x^2 - 9x +15?

Apr 15, 2015

$y = {x}^{3} - 3 {x}^{2} - 9 x + 15$

$y ' = 3 {x}^{2} - 6 x - 9 = 0$ when

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

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

on $\left(- \infty , - 1\right)$ test some number to see that $y '$ is positive
on $\left(- 1 , 3\right)$ test some number to see that $y '$ is negative

So when $x = - 1$, $y$ has a local maximum of $20$
(Put $x = - 1$ in $y = {x}^{3} - 3 {x}^{2} - 9 x + 15$ to get $20$)

on $\left(- 1 , 3\right)$ e already found that $y '$ is negative
on $\left(3 , \infty\right)$ test some number to see that $y '$ is positive

So when $x = 3$, $y$ has a local minimum of $- 12$.
(Put $x = 3$ in $y = {x}^{3} - 3 {x}^{2} - 9 x + 15$ to get $- 12$)