# What are the extrema of f(x)=2x^3-3x^2-36x-3?

Sep 8, 2017

Local maximam is at $\left(- 2 , 41\right)$ and local minimum
is at
$\left(3 , - 84\right)$

#### Explanation:

All local maximums and minimums on a function’s graph are called local extrema.

$f \left(x\right) = 2 {x}^{3} - 3 {x}^{2} - 36 x - 3 \mathmr{and} {f}^{'} \left(x\right) = 6 {x}^{2} - 6 x - 36$

At critical points ${f}^{'} \left(x\right) = 0 \therefore 6 \left({x}^{2} - x - 6\right) = 0$ or

${x}^{2} - x - 6 = 0 \mathmr{and} \left(x - 3\right) \left(x + 2\right) = 0$ ;Critical points are

$x = - 2 , x = 3$ when $x = - 2 , f \left(x\right) = 2 {\left(- 2\right)}^{3} - 3 {\left(- 2\right)}^{2} - 36 \left(- 2\right) - 3 = 41$

when $x = 3 , f \left(x\right) = 2 \cdot {3}^{3} - 3 \cdot {3}^{2} - 36 \cdot 3 - 3 = - 84$ . To test

local maximum or minimum we will examine sign change

at three positions $x = - 3 , x = 0 , x = 4$ for

f^'(x) = 6x^2 -6x- 36 ; f^'( -3) = 36  (increasing) ,

${f}^{'} \left(0\right) = - 36$ (decreasing) and ${f}^{'} \left(4\right) = 36$ (increasing)

$x = - 2$ is local maximum and $x = 3$ is local minimum

Hence local maximam is at $\left(- 2 , 41\right)$ and local minimum

is at $\left(3 , - 84\right)$

graph{2x^3-3x^2-36x-3 [-157, 157, -78.5, 78.5]}