# Find and classify all the turning points of y=2x^3+9x^2-24x-1?

Sep 5, 2017

Maximum $\left(- 4 , 111\right)$

Minimum $\left(1 , - 14\right)$

#### Explanation:

$y = 2 {x}^{3} + 9 {x}^{2} - 24 x - 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 6 {x}^{2} + 18 x - 24$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 12 x + 18$

For critical points need $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$.

$\therefore 0 = 6 {x}^{2} + 18 x - 24$

Divide both sides by 6:

${x}^{2} + 3 x - 4 = 0$

$\left(x + 4\right) \left(x - 1\right) = 0$

$\implies x = - 4 \text{ or } x = 1$

For the nature of these points, need to examine the curvature which is given by the second derivative:

$\textcolor{red}{x = - 4}$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} {|}_{x = - 4} = 12 \left(- 4\right) + 18 = - 30 < 0$

This means that $x = - 4$ is a maximum.

$y \left(- 4\right) = 2 {\left(- 4\right)}^{3} + 9 {\left(- 4\right)}^{2} - 24 \left(- 4\right) - 1$

$y \left(- 4\right) = - 128 + 144 + 96 - 1 = 111$

$\therefore \left(- 4 , 111\right) \text{ is a maximum}$

$\textcolor{red}{x = 1}$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} {|}_{x = 1} = 12 \left(1\right) + 18 = 30 > 0$

This means that $x = 1$ is a minimum.

$y \left(1\right) = 2 {\left(1\right)}^{3} + 9 {\left(1\right)}^{2} - 24 \left(1\right) - 1$

$y \left(1\right) = 2 + 9 - 24 - 1 = - 14$

$\therefore \left(1 , - 14\right) \text{ is a minimum}$