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

The extreme f(x)=(4x)/(x^2+1),-4<= x<= 0 is Maximum(-4,-16/17),Minimum(-1,-2),Maximum(0,0)

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

extreme points

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

Maximum (- 4, - \frac{16}{17}), Minimum (- 1, - 2), Maximum (0, 0)

Solution steps

Find the critical points:

x = - 4, x = - 1, x = 0

Domain of

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

Find intervals:

Decreasing

: - 4 < x < - 1,

Increasing

: - 1 < x < 0

Plug the extreme point

x = - 4

into

\frac{4 x}{x ^{2} + 1} \Rightarrow y = - \frac{16}{17}

Maximum (- 4, - \frac{16}{17})

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 (- 4, - \frac{16}{17}), Minimum (- 1, - 2), Maximum (0, 0)

f (x) = ln (27 + x^{3})

f (x) = x^{3} - 9 x^{2} + 27 x - 3

f (x, y) = y^{2} + x y + 3 x + 2 y + 5

E, E = \frac{I R}{I}

f (x) = 3 + 8 x^{3} - y^{3} - 12 x y

What is the extreme f(x)=(4x)/(x^2+1),-4<= x<= 0 ?
The extreme f(x)=(4x)/(x^2+1),-4<= x<= 0 is Maximum(-4,-16/17),Minimum(-1,-2),Maximum(0,0)