# How do you find the local max and min for f(x)=2x^3 + 5x^2 - 4x - 3?

##### 1 Answer
May 11, 2018

$x = - 2$ is a local maximum, and $x = \frac{1}{3}$ is a local minimum.
See explanations below.

#### Explanation:

f(x)=2x³+5x²-4x-3
What we know is that there's a local extremum when $f ' \left(x\right) = 0$
f'(x)=6x²+10x-4
6x²+10x-4=0
6(x²+5/3x-2/3)=0
x²+5/3x-2/3=0
x²-x/3+2x-2/3=0
$x \left(x - \frac{1}{3}\right) + 2 \left(x - \frac{1}{3}\right) = 0$
$\left(x + 2\right) \left(x - \frac{1}{3}\right) = 0$
Now we can clearly see that when $x = - 2$ and $x = \frac{1}{3}$, $f ' \left(x\right) = 0$
Also, we know that out of roots, f(x)=ax³+bx²+cx+d take the sign of $- a$ in $- \infty$ and the sign of $a$ in $+ \infty$. Because of that, we can deduce that $x = - 2$ is a local maximum, and $x = \frac{1}{3}$ is a local minimum.
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