# How do use the first derivative test to determine the local extrema  f(x) = 3x^4-8x^3-90x^2+50?

Oct 24, 2015

Find the critical numbers and test for min/max using the first derivative test.

#### Explanation:

$f \left(x\right) = 3 {x}^{4} - 8 {x}^{3} - 90 {x}^{2} + 50$

$\text{dom} \left(f\right) = \left(- \infty , \infty\right)$

$f ' \left(x\right) = 12 {x}^{3} - 24 {x}^{2} - 180 x$

$f ' \left(x\right)$ is never undefined

$f ' \left(x\right) = 12 x \left({x}^{2} - 2 x - 15\right) = 12 x \left(x - 5\right) \left(x + 3\right) = 0$ at $x = - 3 , 0 , 5$

Here is one version of the sign chart for $f ' \left(x\right)$

{: (bb"Intervals:",(-oo,-3),(-3,0),(0,5),(5,oo)), (darr bb"Factors"darr,"========","======","=====","======"), (12x, bb" -",bb" -",bb" +",bb" +"), (x-5,bb" -",bb" -",bb" -",bb" +"), (x+3,bb" -",bb" +",bb" +",bb" +"), ("==========","========","======","=====","======"), (bb"Product"=f'(x),bb" -",bb" +",bb" -",bb" +") :}

At $x = - 3$, the sign of $f ' \left(x\right)$ changes from - to +, (so $f$ changes from decreasing to increasing) so $f \left(- 3\right) = - 301$ is a local minimum.

At $x = 0$, the sign of $f ' \left(x\right)$ changes from + to -, (so $f$ changes from increasing to decreasing) so $f \left(0\right) = 50$ is a local maximum.

At $x = 5$, the sign of $f ' \left(x\right)$ changes from - to +, (so $f$ changes from decreasing to increasing) so $f \left(5\right) = - 1325$ is a local minimum.