Question

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

Answer 1

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

Explanation:

`f(x) = 3x^4-8x^3-90x^2+50`

`"dom"(f) = (-oo,oo)`

`f'(x) = 12x^3-24x^2-180x`

`f'(x)` is never undefined

`f'(x) = 12x(x^2-2x-15) = 12x(x-5)(x+3) = 0` at `x= -3, 0, 5`

Here is one version of the sign chart for `f'(x)`

#{: (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'(x)` changes from - to +, (so `f` changes from decreasing to increasing) so `f(-3) = -301` is a local minimum.

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

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