find absolute extreme values of the function on the interval . f(x)=7(x)1/3-5x, [-3,4].
note: The first x is the power of 1/3

Question

find absolute extreme values of the function on the interval . f(x)=7(x)1/3-5x, [-3,4].
note: The first x is the power of 1/3

anonymous · Answer

are you allowed to use calculus?

anonymous · Answer

yes of course. you can use any level of calculus, the education system here in india is a bit ambiguous.

anonymous · Answer

well u only need to know derivatives.

anonymous · Answer

hmmm, i'm very confused about putting the interval values in the function.

anonymous · Answer

if i remember right the  way to find the extremes is by 
f'(x)=0

On your interval  

the [  and ]  mean that -3 and 4 may be included

anonymous · Answer

yeah, so after all that 
$$x ^{2/3}=7\div15$$
$$x ^{2}=(7\div15)^3$$
$$x = \pm7/15\sqrt{7/15}$$
is it correct?