What is the extreme f(x)=ln(x^2+3x+8) ?

The extreme f(x)=ln(x^2+3x+8) is Minimum(-3/2 ,ln(23/4))

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extreme f(x)=ln(x^2+3x+8)

extreme points

f (x) = ln (x^{2} + 3 x + 8)

Minimum (- \frac{3}{2}, ln (\frac{23}{4}))

Solution steps

f^{'} (x) = \frac{2 x + 3}{x ^{2} + 3 x + 8}

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 + 8) : ln (\frac{23}{4})

Minimum (- \frac{3}{2}, ln (\frac{23}{4}))

f (x) = 9 sin (3 x)

f (x, y) = 16 - 8 x + 10 y

f (x) = 20 sin (2 x)

x (3 - x)

f (x) = y \cdot e^{x} + x \cdot ln (y)

What is the extreme f(x)=ln(x^2+3x+8) ?
The extreme f(x)=ln(x^2+3x+8) is Minimum(-3/2 ,ln(23/4))