# How do you find all local maximum and minimum points given y=x^4-2x^2+3?

Oct 28, 2016

$\left(- 1 , 2\right) \mathmr{and} \left(- 1 , 2\right)$ is a local minimum
$\left(0 , 3\right)$ is a local maximum

#### Explanation:

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

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

$4 x = 0 \Rightarrow x = 0$
$x - 1 = 0 \Rightarrow x = 1$
$x + 1 = 0 \Rightarrow x = - 1$

$\textcolor{b r o w n}{y ' < 0}$
$\textcolor{b r o w n}{- \infty < x < - 1 \mathmr{and} 0 < x < 1}$

$\textcolor{b l u e}{y ' > 0}$
$\textcolor{b l u e}{- 1 < x < 0 \mathmr{and} 1 < x < + \infty}$

So,
the function
$y$ decreases for $\textcolor{b r o w n}{- \infty < x < - 1}$ reaching the point $\left(- 1 , 2\right)$
$y$ increases for $\textcolor{b l u e}{- 1 < x < 0}$
So,$\left(\textcolor{red}{- 1 , 2}\right)$ is a local minimum

$y$ increases for $\textcolor{b l u e}{- 1 < x < 0}$ reaching the point$\left(0 , 3\right)$
then it decreases $\textcolor{b r o w n}{0 < x < 1}$
So,$\left(\textcolor{red}{0 , 3}\right)$ is a local maximum

$y$ decreases for $\textcolor{b r o w n}{0 < x < 1}$ reaching the point$\left(1 , 2\right)$then
$y$ increases for $\textcolor{b l u e}{1 < x < \infty}$
So,$\left(\textcolor{red}{- 1 , 2}\right)$ is a local minimum