# How do you find all local maximum and minimum points using the second derivative test given y=x^3-9x^2+24x?

Mar 16, 2017

We have a local maximum at $\left(2 , 20\right)$
We have a local minimum at $\left(4 , 16\right)$

#### Explanation:

Let's calculate the first derivative

$y = {x}^{3} - 9 {x}^{2} + 24 x$

Let $f \left(x\right) = {x}^{3} - 9 {x}^{2} + 24 x$

$f ' \left(x\right) = \frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2} - 18 x + 24$

$= 3 \left({x}^{2} - 6 x + 8\right) = 3 \left(x - 2\right) \left(x - 4\right)$

The critical points are when $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$

That is, when $x = 2$ and $x = 4$

Now, we calculate the second derivative

f''(x)=(d^2y)/dx^2=6x-18#

When $f ' ' \left(x\right) = \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 0$, we have an inflexion point at $x = 3$

Now,

We calculate

$f ' ' \left(2\right) = 12 - 18 = - 6$

As, $f ' ' \left(2\right) < 0$, we have a local maximum at $\left(2 , 20\right)$

$f ' ' \left(4\right) = 24 - 18 = 6$

As, $f ' ' \left(2\right) > 0$, we have a local minimum at $\left(4 , 16\right)$

graph{x^3-9x^2+24x [-36.67, 45.55, -6.95, 34.14]}