What is the extreme f(x)=ln(x^2+7x+14) ?

The extreme f(x)=ln(x^2+7x+14) is Minimum(-7/2 ,ln(7/4))

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extreme f(x)=ln(x^2+7x+14)

extreme points

f (x) = ln (x^{2} + 7 x + 14)

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

Solution steps

f^{'} (x) = \frac{2 x + 7}{x ^{2} + 7 x + 14}

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 + 14) : ln (\frac{7}{4})

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

f (x, y) = y^{3} + x^{2} - 6 x y + 3 x + 6 y - 7

f (x, y) = x^{2} + x y + y^{2} - 3 x - 6 y

f (x) = x^{3} - 5 x^{2} + 7 x

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

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

What is the extreme f(x)=ln(x^2+7x+14) ?
The extreme f(x)=ln(x^2+7x+14) is Minimum(-7/2 ,ln(7/4))