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

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

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

extreme

Maximum (- 1, - 1), Minimum (0, 0)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < - \frac{1}{2},

Decreasing

: - \frac{1}{2} < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 1

into

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

Maximum (- 1, - 1)

Plug

x = 0

into

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

Minimum (0, 0)

Maximum (- 1, - 1), Minimum (0, 0)

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