What is the extreme f(x)=sqrt(x)-2,0<= x<= 4 ?

The extreme f(x)=sqrt(x)-2,0<= x<= 4 is Minimum(0,-2),Maximum(4,0)

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extreme f(x)=sqrt(x)-2,0<= x<= 4

extreme points

f (x) = x - 2, 0 \leq x \leq 4

Minimum (0, - 2), Maximum (4, 0)

Solution steps

Find the critical points:

x = 0, x = 4

Domain of

x - 2, 0 \leq x \leq 4 : 0 \leq x \leq 4

Find intervals:

Increasing

: 0 < x < 4

Plug the extreme point

x = 0

into

x - 2 \Rightarrow y = - 2

Minimum (0, - 2)

Plug the extreme point

x = 4

into

x - 2 \Rightarrow y = 0

Maximum (4, 0)

Minimum (0, - 2), Maximum (4, 0)

y = 7 x - 7 z - 9

y, y < 3 x - 4

f (x) = 2 - x^{4} + 2 x^{2} - y^{2}

f (x) = \frac{ln ( 2 x )}{x}

x^{4} - 6 x^{2} - 8 x - 3

What is the extreme f(x)=sqrt(x)-2,0<= x<= 4 ?
The extreme f(x)=sqrt(x)-2,0<= x<= 4 is Minimum(0,-2),Maximum(4,0)