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)

(- 4, 3), (- 2, - 5)

f (x) = \frac{- 6}{x + 9}

x + 2 y = 6

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

f (x) = - x^{6} + 7 x^{2} + 9 x + 12

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)