What is the extreme f(x)=sqrt(x),0<= x<= 16 ?

The extreme f(x)=sqrt(x),0<= x<= 16 is Minimum(0,0),Maximum(16,4)

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extreme f(x)=sqrt(x),0<= x<= 16

extreme points

f (x) = x, 0 \leq x \leq 16

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

Solution steps

Find the critical points:

x = 0, x = 16

Domain of

x, 0 \leq x \leq 16 : 0 \leq x \leq 16

Find intervals:

Increasing

: 0 < x < 16

Plug the extreme point

x = 0

into

x \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 16

into

x \Rightarrow y = 4

Maximum (16, 4)

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

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

f (x) = 2 x^{3} - 24 x^{2} + 72 x + 10

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

f (x) = x^{3} - x^{2} - 8 x + 8, - 2 \leq x \leq 0

f (x) = \frac{x ^{2} - 11 x + 54}{x - 9}

What is the extreme f(x)=sqrt(x),0<= x<= 16 ?
The extreme f(x)=sqrt(x),0<= x<= 16 is Minimum(0,0),Maximum(16,4)