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

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

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

extreme points

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

Minimum (0, 0)

Solution steps

f^{'} (x) = \frac{3 x ^{2} + 4 x}{2 ( x + 1 ) x + 1}

Find intervals:

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

\frac{x ^{2}}{x + 1} : 0

Minimum (0, 0)

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

f (x) = x^{4} - 2 x^{2} - 8

f (x, y) = 2 - 1 + x + y

f (x) = 3 - 4 sin (3 x), (0, 2 π)

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

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