What is the extreme f(x)=ln(1/(x^2+2x+10)) ?

The extreme f(x)=ln(1/(x^2+2x+10)) is Maximum(-1,-2ln(3))

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extreme f(x)=ln(1/(x^2+2x+10))

extreme points

f (x) = ln (\frac{1}{x ^{2} + 2 x + 10})

Maximum (- 1, - 2 ln (3))

Solution steps

f^{'} (x) = - \frac{2 x + 2}{x ^{2} + 2 x + 10}

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < \infty

Plug

x = - 1

into

ln (\frac{1}{x ^{2} + 2 x + 10}) : - 2 ln (3)

Maximum (- 1, - 2 ln (3))

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

f (x, y) = 72 x + 30 y - 2 x^{2} - 3 y^{2}

y = \frac{( x ^{2} + 1 )}{( x ^{2} - 4 )}

f (x, y) = 3 x^{2} y + y^{3} - 6 x^{2} - 6 y^{2} + 4

f (x, y) = x^{3} + y^{2} - 2 x y + 7 x - 8 y + 4

What is the extreme f(x)=ln(1/(x^2+2x+10)) ?
The extreme f(x)=ln(1/(x^2+2x+10)) is Maximum(-1,-2ln(3))