What is the inflection f(x)=-4/((x^2+1)) ?

The inflection f(x)=-4/((x^2+1)) is (-1/(sqrt(3)),-3),(1/(sqrt(3)),-3)

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inflection f(x)=-4/((x^2+1))

inflection points

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

(- \frac{1}{3}, - 3), (\frac{1}{3}, - 3)

Solution steps

f^{''} (x) = \frac{8 ( - 3 x ^{2} + 1 )}{( x ^{2} + 1 ) ^{3}}

Find intervals:

Concave Downward

: - \infty < x < - \frac{1}{3},

Concave Upward

: - \frac{1}{3} < x < \frac{1}{3},

Concave Downward

: \frac{1}{3} < x < \infty

Plug

x = - \frac{1}{3}

into

- \frac{4}{x ^{2} + 1} : - 3

(- \frac{1}{3}, - 3)

Plug

x = \frac{1}{3}

into

- \frac{4}{x ^{2} + 1} : - 3

(\frac{1}{3}, - 3)

(- \frac{1}{3}, - 3), (\frac{1}{3}, - 3)

What is the inflection f(x)=-4/((x^2+1)) ?
The inflection f(x)=-4/((x^2+1)) is (-1/(sqrt(3)),-3),(1/(sqrt(3)),-3)