What is the extreme f(x)=8x^{3/4}-x,0<= x<= 4096 ?

The extreme f(x)=8x^{3/4}-x,0<= x<= 4096 is Minimum(0,0),Maximum(1296,432),Minimum(4096,0)

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extreme f(x)=8x^{3/4}-x,0<= x<= 4096

extreme points

f (x) = 8 x^{\frac{3}{4}} - x, 0 \leq x \leq 4096

Minimum (0, 0), Maximum (1296, 432), Minimum (4096, 0)

Solution steps

Find the critical points:

x = 0, x = 1296, x = 4096

Domain of

8 x^{\frac{3}{4}} - x, 0 \leq x \leq 4096 : 0 \leq x \leq 4096

Find intervals:

Increasing

: 0 < x < 1296,

Decreasing

: 1296 < x < 4096

Plug the extreme point

x = 0

into

8 x^{\frac{3}{4}} - x \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 1296

into

8 x^{\frac{3}{4}} - x \Rightarrow y = 432

Maximum (1296, 432)

Plug the extreme point

x = 4096

into

8 x^{\frac{3}{4}} - x \Rightarrow y = 0

Minimum (4096, 0)

Minimum (0, 0), Maximum (1296, 432), Minimum (4096, 0)

\frac{x ^{2} - 7}{x + 4}

f (x) = \frac{x ^{3}}{3} + 4 x^{2} + 15 x + 7

T (x, y) = \frac{x ^{3} + y ^{3}}{( x + y ) ^{2} - 3 x y} - x

f (x) = 6 x - 6

What is the extreme f(x)=8x^{3/4}-x,0<= x<= 4096 ?
The extreme f(x)=8x^{3/4}-x,0<= x<= 4096 is Minimum(0,0),Maximum(1296,432),Minimum(4096,0)