What is the extreme f(x)=9x^{2/3}-x,0<= x<= 729 ?

The extreme f(x)=9x^{2/3}-x,0<= x<= 729 is Minimum(0,0),Maximum(216,108),Minimum(729,0)

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extreme f(x)=9x^{2/3}-x,0<= x<= 729

extreme points

f (x) = 9 x^{\frac{2}{3}} - x, 0 \leq x \leq 729

Minimum (0, 0), Maximum (216, 108), Minimum (729, 0)

Solution steps

Find the critical points:

x = 0, x = 216, x = 729

Domain of

9 x^{\frac{2}{3}} - x, 0 \leq x \leq 729 : 0 \leq x \leq 729

Find intervals:

Increasing

: 0 < x < 216,

Decreasing

: 216 < x < 729

Plug the extreme point

x = 0

into

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

Minimum (0, 0)

Plug the extreme point

x = 216

into

9 x^{\frac{2}{3}} - x \Rightarrow y = 108

Maximum (216, 108)

Plug the extreme point

x = 729

into

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

Minimum (729, 0)

Minimum (0, 0), Maximum (216, 108), Minimum (729, 0)

f (x) = \frac{x ^{2} + 4}{4 x}, 1 \leq x \leq 10

23 x + 11 y + 20

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

f (x) = x^{3} - 2 x^{2} - 15 x + 5 (- 2)

f (x) = 4 x^{3} - 16 x + 4

What is the extreme f(x)=9x^{2/3}-x,0<= x<= 729 ?
The extreme f(x)=9x^{2/3}-x,0<= x<= 729 is Minimum(0,0),Maximum(216,108),Minimum(729,0)