What is the inflection-7/(1+x^2) ?

The inflection-7/(1+x^2) is (-1/(sqrt(3)),-21/4),(1/(sqrt(3)),-21/4)

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inflection-7/(1+x^2)

inflection points

- \frac{7}{1 + x ^{2}}

(- \frac{1}{3}, - \frac{21}{4}), (\frac{1}{3}, - \frac{21}{4})

Solution steps

f^{''} (x) = \frac{14 ( - 3 x ^{2} + 1 )}{( 1 + x ^{2} ) ^{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{7}{1 + x ^{2}} : - \frac{21}{4}

(- \frac{1}{3}, - \frac{21}{4})

Plug

x = \frac{1}{3}

into

- \frac{7}{1 + x ^{2}} : - \frac{21}{4}

(\frac{1}{3}, - \frac{21}{4})

(- \frac{1}{3}, - \frac{21}{4}), (\frac{1}{3}, - \frac{21}{4})

f (x) = - \frac{2}{5} x + 3

y = - 29 + 5 (7 - x)

\frac{x - 4}{x + 4}

(5, 1), (1, 1)

f (x) = x^{3} - x^{2} - 8 x

What is the inflection-7/(1+x^2) ?
The inflection-7/(1+x^2) is (-1/(sqrt(3)),-21/4),(1/(sqrt(3)),-21/4)