What is the extreme f(x)=(3x^2)/(1/2 x^2+1) ?

The extreme f(x)=(3x^2)/(1/2 x^2+1) is Minimum(0,0)

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extreme f(x)=(3x^2)/(1/2 x^2+1)

extreme points

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

Minimum (0, 0)

Solution steps

f^{'} (x) = \frac{24 x}{( x ^{2} + 2 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

\frac{3 x ^{2}}{\frac{1}{2} x ^{2} + 1} : 0

Minimum (0, 0)

g (x, y) = x y^{\frac{1}{2}}

f (x) = x^{3} - 6 x^{2} + 12 x - 5

f (x, y) = x^{4} - 4 x^{2} + y^{4} - 8 y^{2} + 5

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

2 sin (x - \frac{π}{6})

What is the extreme f(x)=(3x^2)/(1/2 x^2+1) ?
The extreme f(x)=(3x^2)/(1/2 x^2+1) is Minimum(0,0)