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

The inflection-1/(x^2+4) is (-2/(sqrt(3)),-3/16),(2/(sqrt(3)),-3/16)

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inflection-1/(x^2+4)

inflection points

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

(- \frac{2}{3}, - \frac{3}{16}), (\frac{2}{3}, - \frac{3}{16})

Solution steps

f^{''} (x) = \frac{2 ( - 3 x ^{2} + 4 )}{( x ^{2} + 4 ) ^{3}}

Find intervals:

Concave Downward

: - \infty < x < - \frac{2}{3},

Concave Upward

: - \frac{2}{3} < x < \frac{2}{3},

Concave Downward

: \frac{2}{3} < x < \infty

Plug

x = - \frac{2}{3}

into

- \frac{1}{x ^{2} + 4} : - \frac{3}{16}

(- \frac{2}{3}, - \frac{3}{16})

Plug

x = \frac{2}{3}

into

- \frac{1}{x ^{2} + 4} : - \frac{3}{16}

(\frac{2}{3}, - \frac{3}{16})

(- \frac{2}{3}, - \frac{3}{16}), (\frac{2}{3}, - \frac{3}{16})

f (x) = 12 x^{2} + 2 x^{3}

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

\frac{1 - 4 x}{2 x + 7}

x^{2} - x

f (x) = lo g_{2} (3 - ∣ 2 - x ∣)

What is the inflection-1/(x^2+4) ?
The inflection-1/(x^2+4) is (-2/(sqrt(3)),-3/16),(2/(sqrt(3)),-3/16)