What are the local extrema, if any, of #f(x)= 2x+15x^(2/15)#?

1 Answer
Jul 23, 2017

Answer:

Local maximum of 13 at 1 and local minimum of 0 at 0.

Explanation:

Domain of #f# is #RR#

#f'(x) = 2+2x^(-13/15) = (2x^(13/15)+2)/x^(13/15)#

#f'(x) = 0# at #x = -1# and #f'(x)# does not exist at #x = 0#.

Both #-1# and #9# are in the domain of #f#, so they are both critical numbers.

First Derivative Test:

On #(-oo,-1)#, #f'(x) > 0# (for example at #x = -2^15#)
On #(-1,0)#, #f'(x) < 0# (for example at #x = -1/2^15#)

Therefore #f(-1) = 13# is a local maximum.

On #(0,oo)#, #f'(x) >0# (use any large positive #x#)

So #f(0) = 0# is a local minimum.