What is the extreme f(x)=4x^{1/2}-x,0<= x<= 16 ?

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

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extreme f(x)=4x^{1/2}-x,0<= x<= 16

extreme points

f (x) = 4 x^{\frac{1}{2}} - x, 0 \leq x \leq 16

Minimum (0, 0), Maximum (4, 4), Minimum (16, 0)

Solution steps

Find the critical points:

x = 0, x = 4, x = 16

Domain of

4 x^{\frac{1}{2}} - x, 0 \leq x \leq 16 : 0 \leq x \leq 16

Find intervals:

Increasing

: 0 < x < 4,

Decreasing

: 4 < x < 16

Plug the extreme point

x = 0

into

4 x^{\frac{1}{2}} - x \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 4

into

4 x^{\frac{1}{2}} - x \Rightarrow y = 4

Maximum (4, 4)

Plug the extreme point

x = 16

into

4 x^{\frac{1}{2}} - x \Rightarrow y = 0

Minimum (16, 0)

Minimum (0, 0), Maximum (4, 4), Minimum (16, 0)

- \frac{1}{2} e^{- 2 x} x - \frac{11}{4} e^{- 2 x} + C

f (x) = 2 x e^{- x^{2}}

f (x) = x^{2} \cdot (x - 3)^{2}

- x^{2} + 4

1512 x - 0.14 x^{3}

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