- #1

- 5

- 0

f(x)= 2x + 32/x ; when x > 0

I tried to solve it but I'm not sure if its a maxima or minima, here are my try :

f(x)= 2x + 32/x

f'(x)= 2-32/x^2

2-32/x^2=0

2u^2=32

u^2=16

u=4 or - 4

-4 is not included cause x > 0

f(4)= 2(4) + 32/4

f(4)= 8 + 8

f(4)= 16

(4, 16) but I'm not sure if it an absolute maximum or minimum?

are my steps correct ?