How do use the first derivative test to determine the local extrema #f(x)= x^3 - x^2 - 40x + 8#?

1 Answer
Sep 13, 2015

Answer:

This function has a local maximum value of #2516/27# at #x=-10/3# and a local minimum value of #-104# at #x=4#.

Explanation:

The first derivative is #f'(x)=3x^2-2x-40#. This can be factored as #f'(x)=3x^2-2x-40=(3x+10)(x-4)#. Setting this equal to zero and solving for #x# gives two critical points for #f# at #x=-10/3# and #x=4#.

You can check that #f'# changes sign from positive to negative as #x# increases through #x=-10/3# and negative to positive as #x# increases through #x=4# (note that #f'# is a continuous function and that, for instance, #f'(-4)=48+8-40=16>0#, #f'(0)=-40<0#, and #f'(5)=75-10-40=25>0#).

Therefore, the First Derivative Test implies that #f# has a local maximum at #x=-10/3# and a local minimum at #x=4#.

The local maximum value (output) is #f(-10/3)=-1000/27-100/9+400/3+8=2516/27# and the local minimum value (output) is #f(4)=64-16-160+8=-104#.