What is the extreme f(x)= 2/(sqrt(x^2+4)) ?

The extreme f(x)= 2/(sqrt(x^2+4)) is Maximum(0,1)

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extreme f(x)= 2/(sqrt(x^2+4))

extreme points

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

Maximum (0, 1)

Solution steps

f^{'} (x) = - \frac{2 x}{( x ^{2} + 4 ) ^{\frac{3}{2}}}

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < \infty

Plug

x = 0

into

\frac{2}{x ^{2} + 4} : 1

Maximum (0, 1)

f (x) = 2 x^{2} - x - 6

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

f (x) = x^{4} - 2 x^{2} + 5, - 2 \leq x \leq 2

\frac{x ^{\frac{1}{3}}}{x ^{2} + 8} - 10

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

What is the extreme f(x)= 2/(sqrt(x^2+4)) ?
The extreme f(x)= 2/(sqrt(x^2+4)) is Maximum(0,1)