What is the extreme y=xsqrt(16-x^2) ?

The extreme y=xsqrt(16-x^2) is Maximum(-4,0),Minimum(-2sqrt(2),-8),Maximum(2sqrt(2),8),Minimum(4,0)

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extreme y=xsqrt(16-x^2)

extreme points

y = x 16 - x^{2}

Maximum (- 4, 0), Minimum (- 22, - 8), Maximum (22, 8), Minimum (4, 0)

Solution steps

Find the critical points:

x = - 4, x = - 22, x = 22, x = 4

Domain of

x 16 - x^{2} : - 4 \leq x \leq 4

Find intervals:

Decreasing

: - 4 < x < - 22,

Increasing

: - 22 < x < 22,

Decreasing

: 22 < x < 4

Plug the extreme point

x = - 4

into

x 16 - x^{2} \Rightarrow y = 0

Maximum (- 4, 0)

Plug the extreme point

x = - 22

into

x 16 - x^{2} \Rightarrow y = - 8

Minimum (- 22, - 8)

Plug the extreme point

x = 22

into

x 16 - x^{2} \Rightarrow y = 8

Maximum (22, 8)

Plug the extreme point

x = 4

into

x 16 - x^{2} \Rightarrow y = 0

Minimum (4, 0)

Maximum (- 4, 0), Minimum (- 22, - 8), Maximum (22, 8), Minimum (4, 0)

f (x, y) = 7 e^{y} - 6 y e^{x}

f (x) = \frac{5}{x - 2}

y = x^{2} - 15 x + 54

f (x) = x^{3} - 12 x^{3}

x^{3} + 2 x^{2} + x + 1

What is the extreme y=xsqrt(16-x^2) ?
The extreme y=xsqrt(16-x^2) is Maximum(-4,0),Minimum(-2sqrt(2),-8),Maximum(2sqrt(2),8),Minimum(4,0)