# What are the extrema of y=2x^3 - 5x^2 - 4x + 7?

Apr 13, 2018

$x = - \frac{1}{3}$ local maximum, $x = \frac{5}{6}$ is turning point and $x = 2$ local minimum.

#### Explanation:

After taking derivative of $y$ and equating to $0$,

$6 {x}^{2} - 10 x - 4 = 0$

$\left(3 x + 1\right) \cdot \left(2 x - 4\right) = 0$, so $x 1 = - \frac{1}{3}$ and ${x}_{2} = 2$

For $x = - \frac{1}{3}$, $y = \frac{140}{27}$. So $x = - \frac{1}{3}$ local maximum.

For $x = 2$, $y = - 7$. So $x = 2$ local minimum.

After taking second derivative of $y$ and equating to $0$,

$12 x - 10 = 0$, so $x = \frac{5}{6}$ is turning point.