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

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

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extreme f(x)=x+ln(x^2-1)

extreme points

f (x) = x + ln (x^{2} - 1)

Maximum (- 1 - 2, - 1 - 2 + ln (2 + 22))

Solution steps

f^{'} (x) = 1 + \frac{2 x}{x ^{2} - 1}

Find intervals:

Increasing

: - \infty < x < - 1 - 2,

Decreasing

: - 1 - 2 < x < - 1,

Increasing

: 1 < x < \infty

Plug

x = - 1 - 2

into

x + ln (x^{2} - 1) : - 1 - 2 + ln (2 + 22)

Maximum (- 1 - 2, - 1 - 2 + ln (2 + 22))

f (x) = 2 x^{3} - 33 x^{2} + 168 x + 7

P (x, y) = 10 x^{2} - 10 y^{2}

f (x) = x^{2} e^{1 - x}

(3 - x) (x - 5)^{2}

f (x, y) = \frac{1}{x} + \frac{1}{y} - x y + 3

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