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

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

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

inflection points

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

(- 1, ln (2)), (1, ln (2))

Solution steps

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

Find intervals:

Concave Downward

: - \infty < x < - 1,

Concave Upward

: - 1 < x < 1,

Concave Downward

: 1 < x < \infty

Plug

x = - 1

into

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

(- 1, ln (2))

Plug

x = 1

into

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

(1, ln (2))

(- 1, ln (2)), (1, ln (2))

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