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

1 Answer
Mar 23, 2017

Answer:

#x={-3,0,3}#

Explanation:

Local extrema occur whenever the slope is equal to 0 so we must first find the derivative of the function, set it equal to 0, and then solve for x to find all x's for which there are local extrema.

Using the power-down rule we can find that #f'(x)=8x^3-72x#. Now set it equal to 0. #8x^3-72x=0#. To solve, factor out an #8x# to get #8x(x^2-9)=0# then using the rule of the difference of two squares split #x^2-9# into its two factors to get #8x(x+3)(x-3)=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: #8x=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!