What is the extreme f(x)=(x^2-16)^{1/9},-5<= x<= 4 ?

The extreme f(x)=(x^2-16)^{1/9},-5<= x<= 4 is Maximum(-5,9^{1/9}),Minimum(0,-16^{1/9}),Maximum(4,0)

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extreme f(x)=(x^2-16)^{1/9},-5<= x<= 4

extreme points

f (x) = (x^{2} - 16)^{\frac{1}{9}}, - 5 \leq x \leq 4

Maximum (- 5, 9^{\frac{1}{9}}), Minimum (0, - 1 6^{\frac{1}{9}}), Maximum (4, 0)

Solution steps

Find the critical points:

x = - 5, x = - 4, x = 0, x = 4

Domain of

(x^{2} - 16)^{\frac{1}{9}}, - 5 \leq x \leq 4 : - 5 \leq x \leq 4

Find intervals:

Decreasing

: - 5 < x < - 4,

Decreasing

: - 4 < x < 0,

Increasing

: 0 < x < 4

Plug the extreme point

x = - 5

into

(x^{2} - 16)^{\frac{1}{9}} \Rightarrow y = 9^{\frac{1}{9}}

Maximum (- 5, 9^{\frac{1}{9}})

Plug the extreme point

x = 0

into

(x^{2} - 16)^{\frac{1}{9}} \Rightarrow y = - 1 6^{\frac{1}{9}}

Minimum (0, - 1 6^{\frac{1}{9}})

Plug the extreme point

x = 4

into

(x^{2} - 16)^{\frac{1}{9}} \Rightarrow y = 0

Maximum (4, 0)

Maximum (- 5, 9^{\frac{1}{9}}), Minimum (0, - 1 6^{\frac{1}{9}}), Maximum (4, 0)

y = 6 x - x^{2} + 2

f (x) = \frac{e ^{a x} - e ^{- a x}}{e ^{a x} + e ^{- a x}}

x + ln (x^{2} + 1)

V (B, h) = \frac{B h}{3}

x^{3} + 3 x^{2} + 2 x

What is the extreme f(x)=(x^2-16)^{1/9},-5<= x<= 4 ?
The extreme f(x)=(x^2-16)^{1/9},-5<= x<= 4 is Maximum(-5,9^{1/9}),Minimum(0,-16^{1/9}),Maximum(4,0)