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1 Answer

f'(x)=⅔x^(-⅓)(x-4)+x^⅔=0 at extrema.

That is, ⅔(x-4)+x=0, 2x-8+3x=0 (x≠0), 5x=8, x=8/5=1.6.

f(x)=-3.283 approx.

f(0)=0, f(4)=0.

f(-5)=-26.316 approx. f(5)=2.924approx.

(1) x=0, x=8/5.

(2) Local max at x=0, f(0)=0; Local min at x=8/5 and f(8/5)=-3.28.

(3) Abs max at x=5 and f(5)=2.92; abs min at x=-5 and f(-5)=-26.32.

by Top Rated User (797k points)

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