What is the extreme f(x)=-(40x)/(x^2+25) ?

The extreme f(x)=-(40x)/(x^2+25) is Maximum(-5,4),Minimum(5,-4)

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extreme f(x)=-(40x)/(x^2+25)

extreme points

f (x) = - \frac{40 x}{x ^{2} + 25}

Maximum (- 5, 4), Minimum (5, - 4)

Solution steps

f^{'} (x) = - \frac{40 ( - x ^{2} + 25 )}{( x ^{2} + 25 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 5,

Decreasing

: - 5 < x < 5,

Increasing

: 5 < x < \infty

Plug

x = - 5

into

- \frac{40 x}{x ^{2} + 25} : 4

Maximum (- 5, 4)

Plug

x = 5

into

- \frac{40 x}{x ^{2} + 25} : - 4

Minimum (5, - 4)

Maximum (- 5, 4), Minimum (5, - 4)

f (x) = 2 x^{2} - 12 x + 1

g (x) = 3 - x

2 + 24.5 x - 4.9 x^{2}

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

x + \frac{4}{x}

What is the extreme f(x)=-(40x)/(x^2+25) ?
The extreme f(x)=-(40x)/(x^2+25) is Maximum(-5,4),Minimum(5,-4)