What is the extreme f(x)=(10x)/(3x^2+3) ?

The extreme f(x)=(10x)/(3x^2+3) is Minimum(-1,-5/3),Maximum(1, 5/3)

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extreme f(x)=(10x)/(3x^2+3)

extreme points

f (x) = \frac{10 x}{3 x ^{2} + 3}

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

Solution steps

f^{'} (x) = - \frac{10 ( x ^{2} - 1 )}{3 ( x ^{2} + 1 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - 1

into

\frac{10 x}{3 x ^{2} + 3} : - \frac{5}{3}

Minimum (- 1, - \frac{5}{3})

Plug

x = 1

into

\frac{10 x}{3 x ^{2} + 3} : \frac{5}{3}

Maximum (1, \frac{5}{3})

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

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

3 x^{4} + 7 x^{3} - 1 8^{2} - 32

f (x) = 0.2 x^{2} - 20 x + 900

14 + 4 x - x^{2}

f (x) = - x (x + 1)

What is the extreme f(x)=(10x)/(3x^2+3) ?
The extreme f(x)=(10x)/(3x^2+3) is Minimum(-1,-5/3),Maximum(1, 5/3)