What is the extreme-(8*x)/(x^2+1) ?

The extreme-(8*x)/(x^2+1) is Maximum(-1,4),Minimum(1,-4)

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extreme-(8*x)/(x^2+1)

extreme points

- \frac{8 \cdot x}{x ^{2} + 1}

Maximum (- 1, 4), Minimum (1, - 4)

Solution steps

f^{'} (x) = - \frac{8 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 1

into

- \frac{8 \cdot x}{x ^{2} + 1} : 4

Maximum (- 1, 4)

Plug

x = 1

into

- \frac{8 \cdot x}{x ^{2} + 1} : - 4

Minimum (1, - 4)

Maximum (- 1, 4), Minimum (1, - 4)

\frac{x ^{2}}{x ^{2} - 36}

f (x) = - e^{x} + 15 e^{\frac{x}{14}}

f (x) = x^{2} - y^{2} - 5 x - 4 y

20 q - 2 q^{2}

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

What is the extreme-(8*x)/(x^2+1) ?
The extreme-(8*x)/(x^2+1) is Maximum(-1,4),Minimum(1,-4)