# How do you find the local maximum and minimum values of f(x)=2x^3 + 5x^2 - 4x - 3?

Jun 10, 2017

$f \left(- 2\right) = 9$ is local Maxima, $f \left(\frac{1}{3}\right) = - \frac{100}{27}$ is local

Minima.

#### Explanation:

We know that, for local Extreme values, $f ' \left(x\right) = 0.$

Also, $f ' ' \left(x\right) < 0$ for Maxima, and, $f ' ' \left(x\right) > 0$ for Minima.

$f \left(x\right) = 2 {x}^{3} + 5 {x}^{2} - 4 x - 3$

 rArr f'(x)=6x^2+10x-4, &, f''(x)=12x+10.

$f ' \left(x\right) = 0 \Rightarrow 2 \left(3 {x}^{2} + 5 x - 2\right) = 0.$

$\Rightarrow 2 \left(x + 2\right) \left(3 x - 1\right) = 0.$

$\Rightarrow x = - 2 , x = \frac{1}{3.}$

Now, $f ' ' \left(- 2\right) = - 24 + 10 = - 14 < 0.$

$\therefore f$ has a local maxima at $x = - 2 ,$ &, it is,

$f \left(- 2\right) = - 16 + 20 + 8 - 3 = 9.$

Also, $f ' ' \left(\frac{1}{3}\right) = 4 + 10 = 14 > 0.$

$\therefore f$ has a local minima at $x = \frac{1}{3} ,$ which is,

$f \left(\frac{1}{3}\right) = \frac{2}{27} + \frac{5}{9} - \frac{4}{3} - 3 = - \frac{100}{27.}$

Thus, $f \left(- 2\right) = 9$ is local Maxima, $f \left(\frac{1}{3}\right) = - \frac{100}{27}$ is local

Minima.

Enjoy Maths.!