What is the inflection f(x)=x^4-8x^2+1 ?

The inflection f(x)=x^4-8x^2+1 is (-2/(sqrt(3)),-71/9),(2/(sqrt(3)),-71/9)

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inflection f(x)=x^4-8x^2+1

inflection points

f (x) = x^{4} - 8 x^{2} + 1

(- \frac{2}{3}, - \frac{71}{9}), (\frac{2}{3}, - \frac{71}{9})

Solution steps

f^{''} (x) = 12 x^{2} - 16

Find intervals:

Concave Upward

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

Concave Downward

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

Concave Upward

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

Plug

x = - \frac{2}{3}

into

x^{4} - 8 x^{2} + 1 : - \frac{71}{9}

(- \frac{2}{3}, - \frac{71}{9})

Plug

x = \frac{2}{3}

into

x^{4} - 8 x^{2} + 1 : - \frac{71}{9}

(\frac{2}{3}, - \frac{71}{9})

(- \frac{2}{3}, - \frac{71}{9}), (\frac{2}{3}, - \frac{71}{9})

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

\frac{2 x - 1}{3} + 1

x^{\frac{1}{x}}

f (x) = (\frac{1}{2})^{2 x} + 5

y = - 3 x + 8

What is the inflection f(x)=x^4-8x^2+1 ?
The inflection f(x)=x^4-8x^2+1 is (-2/(sqrt(3)),-71/9),(2/(sqrt(3)),-71/9)