What is the extreme f(x)=ln(x^2+7x+15) ?

The extreme f(x)=ln(x^2+7x+15) is Minimum(-7/2 ,ln(11/4))

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extreme f(x)=ln(x^2+7x+15)

extreme points

f (x) = ln (x^{2} + 7 x + 15)

Minimum (- \frac{7}{2}, ln (\frac{11}{4}))

Solution steps

f^{'} (x) = \frac{2 x + 7}{x ^{2} + 7 x + 15}

Find intervals:

Decreasing

: - \infty < x < - \frac{7}{2},

Increasing

: - \frac{7}{2} < x < \infty

Plug

x = - \frac{7}{2}

into

ln (x^{2} + 7 x + 15) : ln (\frac{11}{4})

Minimum (- \frac{7}{2}, ln (\frac{11}{4}))

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

x^{3} - 3 x^{2} + 5

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

f (x, y) = \frac{1}{5} x^{5} - \frac{5}{3} x^{3} + 4 x + x y^{2}

f (x) = x^{4} - 12 x^{2} + 36

What is the extreme f(x)=ln(x^2+7x+15) ?
The extreme f(x)=ln(x^2+7x+15) is Minimum(-7/2 ,ln(11/4))