What is the extreme f(x)=ln(x^2+3x+5) ?

The extreme f(x)=ln(x^2+3x+5) is Minimum(-3/2 ,ln(11/4))

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extreme f(x)=ln(x^2+3x+5)

extreme points

f (x) = ln (x^{2} + 3 x + 5)

Minimum (- \frac{3}{2}, ln (\frac{11}{4}))

Solution steps

f^{'} (x) = \frac{2 x + 3}{x ^{2} + 3 x + 5}

Find intervals:

Decreasing

: - \infty < x < - \frac{3}{2},

Increasing

: - \frac{3}{2} < x < \infty

Plug

x = - \frac{3}{2}

into

ln (x^{2} + 3 x + 5) : ln (\frac{11}{4})

Minimum (- \frac{3}{2}, ln (\frac{11}{4}))

f (x) = \frac{x ^{7}}{e ^{4 x}}

25 - x^{2} - y^{2}

y = \frac{3 x - 6}{x + 2}

y = 5 - x^{2}

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

What is the extreme f(x)=ln(x^2+3x+5) ?
The extreme f(x)=ln(x^2+3x+5) is Minimum(-3/2 ,ln(11/4))