# What are the local extrema of f(x)= x^3-3x^2-9x+7?

Feb 11, 2017

Local maximum : $f \left(- 1\right) = 12$. Local minimum :$f \left(3\right) = - 20$.

#### Explanation:

$f = {x}^{3} \left(1 - \frac{3}{x} - \frac{9}{x} ^ 2 + \frac{7}{x} ^ 3\right) \to \pm \infty$, as $x \to \pm \infty$.

f'=3(x^2-2x-3)=0, at x = -1 and 3.

$f ' ' = 6 x - 6 , < 9$, at $x = - 1 , > 0$, at $x = 3 \mathmr{and} = 0$, at $x = 1.$

So, $l o c a l - \max f = f \left(- 1\right) = 12 \mathmr{and} l o c a l - \min f = f \left(3\right) = - 20$.

As, $f ' ' ' \ne 0 , \left(1 , - 4\right)$ is a POI ( point of inflexion ).

graph{(x^3-3x^2-9x+7-y)((x-1)^2+(y+4)^2-.01)=0 [-34, 34, -21, 13]}