What is the extreme f(x)=sqrt(64-x^2),-8<= x<= 8 ?

The extreme f(x)=sqrt(64-x^2),-8<= x<= 8 is Minimum(-8,0),Maximum(0,8),Minimum(8,0)

Popular >

extreme f(x)=sqrt(64-x^2),-8<= x<= 8

extreme points

f (x) = 64 - x^{2}, - 8 \leq x \leq 8

Minimum (- 8, 0), Maximum (0, 8), Minimum (8, 0)

Solution steps

Find the critical points:

x = - 8, x = 0, x = 8

Domain of

64 - x^{2}, - 8 \leq x \leq 8 : - 8 \leq x \leq 8

Find intervals:

Increasing

: - 8 < x < 0,

Decreasing

: 0 < x < 8

Plug the extreme point

x = - 8

into

64 - x^{2} \Rightarrow y = 0

Minimum (- 8, 0)

Plug the extreme point

x = 0

into

64 - x^{2} \Rightarrow y = 8

Maximum (0, 8)

Plug the extreme point

x = 8

into

64 - x^{2} \Rightarrow y = 0

Minimum (8, 0)

Minimum (- 8, 0), Maximum (0, 8), Minimum (8, 0)

- \frac{2 e ^{x}}{- 4 x - 1}

f (x) = \frac{4}{4 x ^{2} - 1} + b

f (x) = 3 x^{2} - 36 x + 81

K (r, s) = 3 r - 4 s

f (9) = \frac{t}{t - 4}

What is the extreme f(x)=sqrt(64-x^2),-8<= x<= 8 ?
The extreme f(x)=sqrt(64-x^2),-8<= x<= 8 is Minimum(-8,0),Maximum(0,8),Minimum(8,0)