What is the extreme f(x)=(4x)/(x^2+1),-3<= x<= 0 ?

The extreme f(x)=(4x)/(x^2+1),-3<= x<= 0 is Maximum(-3,-6/5),Minimum(-1,-2),Maximum(0,0)

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extreme f(x)=(4x)/(x^2+1),-3<= x<= 0

extreme points

f (x) = \frac{4 x}{x ^{2} + 1}, - 3 \leq x \leq 0

Maximum (- 3, - \frac{6}{5}), Minimum (- 1, - 2), Maximum (0, 0)

Solution steps

Find the critical points:

x = - 3, x = - 1, x = 0

Domain of

\frac{4 x}{x ^{2} + 1}, - 3 \leq x \leq 0 : - 3 \leq x \leq 0

Find intervals:

Decreasing

: - 3 < x < - 1,

Increasing

: - 1 < x < 0

Plug the extreme point

x = - 3

into

\frac{4 x}{x ^{2} + 1} \Rightarrow y = - \frac{6}{5}

Maximum (- 3, - \frac{6}{5})

Plug the extreme point

x = - 1

into

\frac{4 x}{x ^{2} + 1} \Rightarrow y = - 2

Minimum (- 1, - 2)

Plug the extreme point

x = 0

into

\frac{4 x}{x ^{2} + 1} \Rightarrow y = 0

Maximum (0, 0)

Maximum (- 3, - \frac{6}{5}), Minimum (- 1, - 2), Maximum (0, 0)

f (x) = \frac{4 x + 9}{x ^{2} + 9}

f (x, y) = 2 x y - 1

f (2.4) = 3 - 6 (x - 2) + 9 (y - 4)

- 4 x^{4} - 9 x^{3} + 2 x^{2} + 7 x - 2

f (x) = e^{- 4 x - 4 y}

What is the extreme f(x)=(4x)/(x^2+1),-3<= x<= 0 ?
The extreme f(x)=(4x)/(x^2+1),-3<= x<= 0 is Maximum(-3,-6/5),Minimum(-1,-2),Maximum(0,0)