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

1 Answer
Oct 24, 2015

Answer:

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.