What is the extreme f(x)=(x+2)^{4/3},-6<= x<= 6 ?

The extreme f(x)=(x+2)^{4/3},-6<= x<= 6 is Maximum(-6,4*4^{1/3}),Minimum(-2,0),Maximum(6,16)

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extreme f(x)=(x+2)^{4/3},-6<= x<= 6

extreme points

f (x) = (x + 2)^{\frac{4}{3}}, - 6 \leq x \leq 6

Maximum (- 6, 4 \cdot 4^{\frac{1}{3}}), Minimum (- 2, 0), Maximum (6, 16)

Solution steps

Find the critical points:

x = - 6, x = - 2, x = 6

Domain of

(x + 2)^{\frac{4}{3}}, - 6 \leq x \leq 6 : - 6 \leq x \leq 6

Find intervals:

Decreasing

: - 6 < x < - 2,

Increasing

: - 2 < x < 6

Plug the extreme point

x = - 6

into

(x + 2)^{\frac{4}{3}} \Rightarrow y = 4 4^{\frac{1}{3}}

Maximum (- 6, 4 \cdot 4^{\frac{1}{3}})

Plug the extreme point

x = - 2

into

(x + 2)^{\frac{4}{3}} \Rightarrow y = 0

Minimum (- 2, 0)

Plug the extreme point

x = 6

into

(x + 2)^{\frac{4}{3}} \Rightarrow y = 16

Maximum (6, 16)

Maximum (- 6, 4 \cdot 4^{\frac{1}{3}}), Minimum (- 2, 0), Maximum (6, 16)

f (x) = - x^{3} + 15 x^{2}

f (x) = \frac{x ^{2} - 14 x + 49}{x - 10}

f (x) = e^{x^{3} - 12 x}

(x - 8 x^{2})^{\frac{1}{3}}

f (x) = x 4 - x^{2}, - 2 \leq x \leq 2

What is the extreme f(x)=(x+2)^{4/3},-6<= x<= 6 ?
The extreme f(x)=(x+2)^{4/3},-6<= x<= 6 is Maximum(-6,4*4^{1/3}),Minimum(-2,0),Maximum(6,16)