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

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

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

extreme points

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

Minimum (- \frac{1}{2}, \frac{1}{2})

Solution steps

f^{'} (x) = \frac{6 x + 3}{2 3 x ^{2} + 3 x + 1}

Find intervals:

Decreasing

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

Increasing

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

Plug

x = - \frac{1}{2}

into

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

Minimum (- \frac{1}{2}, \frac{1}{2})

x^{2} + x y + y^{2} - 2 x - y

f (x) = x^{3} + y^{2} + 2 x y - 4 x - 3 y + 5

f (x, y) = 2 x^{2} + 3 x y + 4 y^{2} - 7 x - 11 y

f (x) = \frac{4 x}{x ^{2} + 9}

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

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