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

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

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

extreme points

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

Saddle (- 1, - 1 + ln (2))

Solution steps

Find the critical points:

x = - 1

Domain of

x + ln (x^{2} + 1) : - \infty < x < \infty

Find intervals:

Increasing

: - \infty < x < - 1,

Increasing

: - 1 < x < \infty

Plug the extreme point

x = - 1

into

x + ln (x^{2} + 1) \Rightarrow y = - 1 + ln (2)

Saddle (- 1, - 1 + ln (2))

f (x, y) = \frac{x ^{2} + y ^{2} - 16}{x - 3}

f (x) = x^{2} + x y + y^{2} - 12 x + 1

f (x) = x^{2} - 8 x + 10

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

f (x) = x^{2} + 9 x - 5

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