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))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

extreme

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}))

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))