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

The extreme f(x)=ln(x^2+x+1) is Minimum(-1/2 ,ln(3/4))

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

extreme points

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

Minimum (- \frac{1}{2}, ln (\frac{3}{4}))

Solution steps

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

Find intervals:

Decreasing

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

Increasing

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

Plug

x = - \frac{1}{2}

into

ln (x^{2} + x + 1) : ln (\frac{3}{4})

Minimum (- \frac{1}{2}, ln (\frac{3}{4}))

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

V, V = 13 π r 2 h

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

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

f (x) = x x - 2

What is the extreme f(x)=ln(x^2+x+1) ?
The extreme f(x)=ln(x^2+x+1) is Minimum(-1/2 ,ln(3/4))