What is the extreme f(x)= 1/(sqrt(x^2+1)) ?

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

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extreme f(x)= 1/(sqrt(x^2+1))

extreme points

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

Maximum (0, 1)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < \infty

Plug

x = 0

into

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

Maximum (0, 1)

x^{2} + \frac{405000}{x}

\frac{2 x ^{2} - 14 x + 24}{x ^{2} + 6 x + 40}

f (x) = x (5 - x) (8 - x), 0 < x < 5

f (x) = \frac{( x ^{2} + 121 )}{2 x}

f (x) = - 7 + 12 x - 3 x^{2}

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