What is the extreme f(x)=ln(x^2+5x+9) ?

The extreme f(x)=ln(x^2+5x+9) is Minimum(-5/2 ,ln(11/4))

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extreme f(x)=ln(x^2+5x+9)

extreme points

f (x) = ln (x^{2} + 5 x + 9)

Minimum (- \frac{5}{2}, ln (\frac{11}{4}))

Solution steps

f^{'} (x) = \frac{2 x + 5}{x ^{2} + 5 x + 9}

Find intervals:

Decreasing

: - \infty < x < - \frac{5}{2},

Increasing

: - \frac{5}{2} < x < \infty

Plug

x = - \frac{5}{2}

into

ln (x^{2} + 5 x + 9) : ln (\frac{11}{4})

Minimum (- \frac{5}{2}, ln (\frac{11}{4}))

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

f (χ) = χ^{2} e^{- 4 χ^{2}}

f (x) = \frac{y - 800}{- 2}

(x - 1) e^{\frac{1}{x}}

f (x) = x^{4} + x^{2} (y - 2) + 3 (y - 1)^{2}

What is the extreme f(x)=ln(x^2+5x+9) ?
The extreme f(x)=ln(x^2+5x+9) is Minimum(-5/2 ,ln(11/4))