A differentiable function #f# has only one critical number: #x=-3#. Identify the relative extrema of #f# at (-3, f(-3)) if #f'(-4)=(1/2)# and #f'(-2)=-1#?

1 Answer
Apr 20, 2017

Answer:

#f(-3)# is a relative maximum.

Explanation:

#f'(-3)# is either undefined or #0#.

Since #f# is differentiable, #f'# is never undefined. That, together with "#-3# is a critical number", implies that #f'(-3) = 0#.

Derivative have the intermediate value property.

Therefore, since #f'(-4) > 0# , there is nowhere in #[-4,0)# where #f'# is #0#, we can conclude that #f'(x) > 0# for #x in #[-4,-3)# so #f# is increasing on #[-4,-3)#.

Similarly, since #f'(-2) < 0# , we can conclude that #f'(x) < 0# for #x in #(-3,-2]# so #f# is decreasing on #(-3,-2]#.

Therefore #f(-3)# is a relative maximum.