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

1 Answer
Sep 15, 2015

Answer:

See the explanation.

Explanation:

Find the critical numbers for #f#.

#f(x)=x^3-2x +pi #

#f'(x)=3x^2-2 #

#f'# is never undefined and is #f'(x) = 0# at #x= +- sqrt6/3#

On #(-oo, -sqrt6/3)#, we get #f'(x)# is positive
on #(-sqrt6/3, sqrt6/3)#, we get #f'(x)# is negative.

So #f# changes from increasing to decreasing as we move to the left past #x = -sqrt6/3#.
Therefore, #f(-sqrt6/3)# is a local maximum.

Recall: on #(-sqrt6/3, sqrt6/3)#, we get #f'(x)# is negative.
now, on #(sqrt6/3, oo)#, we get #f'(x)# is positive.
So #f# changes from decreasing to increasing as we move to the left past #x = sqrt6/3#.
Therefore, #f(sqrt6/3)# is a local minimum.

Calculating (simplifying) #f(-sqrt6/3)# and #f(sqrt6/3)# is left to the student.