What is the extreme f(x)=3x^{2/3}-x,0<= x<= 27 ?

The extreme f(x)=3x^{2/3}-x,0<= x<= 27 is Minimum(0,0),Maximum(8,4),Minimum(27,0)

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extreme f(x)=3x^{2/3}-x,0<= x<= 27

extreme points

f (x) = 3 x^{\frac{2}{3}} - x, 0 \leq x \leq 27

Minimum (0, 0), Maximum (8, 4), Minimum (27, 0)

Solution steps

Find the critical points:

x = 0, x = 8, x = 27

Domain of

3 x^{\frac{2}{3}} - x, 0 \leq x \leq 27 : 0 \leq x \leq 27

Find intervals:

Increasing

: 0 < x < 8,

Decreasing

: 8 < x < 27

Plug the extreme point

x = 0

into

3 x^{\frac{2}{3}} - x \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 8

into

3 x^{\frac{2}{3}} - x \Rightarrow y = 4

Maximum (8, 4)

Plug the extreme point

x = 27

into

3 x^{\frac{2}{3}} - x \Rightarrow y = 0

Minimum (27, 0)

Minimum (0, 0), Maximum (8, 4), Minimum (27, 0)

f (x) = \frac{5 x}{x ^{2} + 9}

x^{9} - 9 x

x^{2} + 4 x - 5

f (x) = 2 sin (x) - x, 0 \leq x \leq 2 π

f (x) = 8 + 6 x - 7 x^{2} - 4 x^{3}

What is the extreme f(x)=3x^{2/3}-x,0<= x<= 27 ?
The extreme f(x)=3x^{2/3}-x,0<= x<= 27 is Minimum(0,0),Maximum(8,4),Minimum(27,0)