# What are the local extrema, if any, of f (x) = 2x^4-36x^2+5 ?

Mar 23, 2017

$x = \left\{- 3 , 0 , 3\right\}$
Using the power-down rule we can find that $f ' \left(x\right) = 8 {x}^{3} - 72 x$. Now set it equal to 0. $8 {x}^{3} - 72 x = 0$. To solve, factor out an $8 x$ to get $8 x \left({x}^{2} - 9\right) = 0$ then using the rule of the difference of two squares split ${x}^{2} - 9$ into its two factors to get $8 x \left(x + 3\right) \left(x - 3\right) = 0$. Now set each of these separately equal to 0 because the entire expression will be 0 when any of the terms are 0.
This gives you 3 equations: $8 x = 0$, $x + 3 = 0$, and $x - 3 = 0$. To solve the first one divide both sides by 8 to get $x = 0$. For the second, subtract 3 from both sides to get $x = - 3$. Lastly, for the third, add 3 to both sides to get $x = 3$. These are all the x-values where local extrema will occur. Hope I helped!