How do you graph f(x)=x^3-3x^2-9x+6 using the information given by the first derivative?

Nov 2, 2016

$f \left(x\right)$ has a local maximun at $\left(- 1 , 11\right)$ and a local minimum at $\left(3 , - 21\right)$

Explanation:

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

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

$f \left(x\right)$ will have turning points where $f ' \left(x\right) = 0$

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

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

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

$\therefore f \left(x\right)$ has turning points at $x = 3$ and $x = - 1$

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

$f ' ' \left(3\right) = 12 > 0 \to x = 3$ is a local minimum
$f ' ' \left(- 1\right) = - 12 < 0 \to x = - 1$ is a local maximum

These points can be seen on the graph of $f \left(x\right)$ below:

graph{x^3-3x^2-9x+6 [-43.74, 46.25, -24.88, 20.13]}