Absolute Extrema

Question

Find the absolute extrema of f(x) = x^2/3 (3-x) on [-2, 2]

Answer 1

f(x) = x^2/3(3-x)

f'(x) = (2/3)x^-1/3(3-x) - x^2/3 = 2(3-x) / (3x^1/3) - x^2/3 = [2(3-x) - 3x] / (3x^1/3) = (6 - 5x) / (3x^1/3)

f'(x) = 0 when x = 6/5 and f'(x) is undefined when x = 0

Evaluate f(x) at 6/5, 0, and at each endpoint.

f(0) = 0

f(6/5) = 2.0326

f(2) = 1.5874

f(-2) = 7.9370

Absolute maximum value on the interval [-2,2] is 7.9370

Absolute minimum value on the interval [-2,2] is 0

Answer 2

ASSUMING you mean f(x) = x^2 / [ 3 (3-x)] = x^2 / (9 - 3x)

x=3 is not in the domain.

x=0 is a solution

f(-2) = 4/ (3* (3 - -2)) = 4/15

f(2) = 4 / (3 * (3-2)) = 4/ (3 * 1) = 4/3

Derivative by quotient rule is f'(x) = [(9-3x)(2x) - (x^2)(-3)]/ (9-3x)^2

= [ 18x - 6x^2 + 3x^2]/(9-3x)^2

= [18x - 3x^2]/(9-3x)^2 = [18x - 3x^2]/(3(3-x))^2

= (3x)(6 - x)/ [ 9(3-x)^2]

= x(6-x)/[3(3-x)^2]

there are 2 critical values at x=0 and x=6, the latter of which is not in the domain.

So (0,0) pulls double duty as a solution and a critical point. (possibly a flex point too, if x appears

as a factor in the 2nd derivative)

sign table shows for x=-1, f(-1) = 1/12 and f'(-1) <0 which means the function is positive but decreasing.

for x = 1, f(1) = 1/6 and f'(1) = 5/12>0 which means the function is positive and increasing.

the absolute max is therefore at the endpoint (2, 4/3)

the absolute min is (0,0)