What is the extreme f(x)=x^{8/3},-8<= x<= 1 ?

The extreme f(x)=x^{8/3},-8<= x<= 1 is Maximum(-8,256),Minimum(0,0),Maximum(1,1)

extreme f(x)=x^{8/3},-8<= x<= 1

extreme points

f (x) = x^{\frac{8}{3}}, - 8 \leq x \leq 1

Maximum (- 8, 256), Minimum (0, 0), Maximum (1, 1)

Solution steps

Find the critical points:

x = - 8, x = 0, x = 1

Domain of

x^{\frac{8}{3}}, - 8 \leq x \leq 1 : - 8 \leq x \leq 1

Find intervals:

Decreasing

: - 8 < x < 0,

Increasing

: 0 < x < 1

Plug the extreme point

x = - 8

into

x^{\frac{8}{3}} \Rightarrow y = 256

Maximum (- 8, 256)

Plug the extreme point

x = 0

into

x^{\frac{8}{3}} \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 1

into

x^{\frac{8}{3}} \Rightarrow y = 1

Maximum (1, 1)

Maximum (- 8, 256), Minimum (0, 0), Maximum (1, 1)

f (x) = cos^{2} (x), 0 \leq x \leq π

f (x) = x^{3} + 2 x^{2} - 7 x

f (x, y) = x e^{- 3 x^{2} - 3 y^{2}}

f (x, y) = 2 x^{2} - 8 x + y^{2} - 8 y + 2

\frac{1}{x - 2}

What is the extreme f(x)=x^{8/3},-8<= x<= 1 ?
The extreme f(x)=x^{8/3},-8<= x<= 1 is Maximum(-8,256),Minimum(0,0),Maximum(1,1)