Verified answer

take the derivative { trivial } , find the critical values ; take the 2nd derivative , evaluate this at

the critical values; if > 0 then local min ; if < 0 then local max ; if 0 then look for regions of

increase and regions of decrease and apply logic...both max and min exist

Differentiate.

5 - 5x^(-1/3) = 0

1 - x^(-1/3) = 0

x^(-1/3) = 1

x^(1/3) = 1 ; x = 1 is the only solution on the reals.

Let's find if it is min or max. Second derivative.

(5/3)x^(-4/3) for all x must be positive, so it is a relative minimum.

5(1) - 15(1)^(1/3) = 5 - 15 = -10.

So the point (1, -10) is a relative minimum.