What is the extreme f(x)=2xsqrt(4-x^2) ?

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

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extreme f(x)=2xsqrt(4-x^2)

extreme points

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

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

Solution steps

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

Find intervals:

Decreasing

: - 2 < x < - 2,

Increasing

: - 2 < x < 2,

Decreasing

: 2 < x < 2

Plug

x = - 2

into

2 x 4 - x^{2} : 0

Maximum (- 2, 0)

Plug

x = - 2

into

2 x 4 - x^{2} : - 4

Minimum (- 2, - 4)

Plug

x = 2

into

2 x 4 - x^{2} : 4

Maximum (2, 4)

Plug

x = 2

into

2 x 4 - x^{2} : 0

Minimum (2, 0)

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

f (x) = 9 x^{2} - x^{3} - 3

f (x) = - x

y = \frac{1}{x + 5}

9 x - 7 y = - 7

x^{4} - 32 x^{3} + 290 x^{2} - 800 x + 6625

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