[Calculus: Absolute Maxima and Minima] Find the maximum and minimum values of the function f(x) = x^19/3 − 10x ^13/3 on the interval [ − 3 , 3 ]

Question

I've been doing pretty good on these problems, however I have not been able to solve this one.

Find the maximum and minimum values of the function ^19/313/3 on the interval .

Minimum value =    Maximum value =   

Answer 1

The function f(x) is an odd function: f(-x) = -f(x), f(0) = 0.

Find the first derivative:

f '(x) = 19/3 x^16/3 - 130/3 x^10/3 = 1/3 x^10/3 (19x² - 130)

f '(x) = 0 if x = 0 or if 19x² - 130 = 0

x = ±√(130/19) ≈ ±2.6157

For the interval [0,3]:

(1) f '(x) < 0 if 0 < x < √(130/19)

f(x) is decreasing. Evaluate: f(√(130/19)) ≈ -203.69

(2) f '(x) > 0 if √(130/19) < x < 3

f(x) is increasing. Evaluate: f(3) ≈ -116.82

Because f(x) is an odd function, then on the interval [-3,0]:

(1) on the interval [-3, - √(130/19)] f(x) is increasing from 116.82 to 203.69

(2) on the interval [ - √(130/19), 0] f(x) is decreasing from 203.69 to 0.

Finally, on the interval [-3,3]

Minimum value ≈ -203.69   Maximum value ≈ 203.69

Exact value for the maximum is 10(130/19)^13/6 - (130/19)^19/6