What is the inflection f(x)=x^4-32x^2+1 ?

The inflection f(x)=x^4-32x^2+1 is (-4/(sqrt(3)),-1271/9),(4/(sqrt(3)),-1271/9)

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inflection f(x)=x^4-32x^2+1

inflection points

f (x) = x^{4} - 32 x^{2} + 1

(- \frac{4}{3}, - \frac{1271}{9}), (\frac{4}{3}, - \frac{1271}{9})

Solution steps

f^{''} (x) = 12 x^{2} - 64

Find intervals:

Concave Upward

: - \infty < x < - \frac{4}{3},

Concave Downward

: - \frac{4}{3} < x < \frac{4}{3},

Concave Upward

: \frac{4}{3} < x < \infty

Plug

x = - \frac{4}{3}

into

x^{4} - 32 x^{2} + 1 : - \frac{1271}{9}

(- \frac{4}{3}, - \frac{1271}{9})

Plug

x = \frac{4}{3}

into

x^{4} - 32 x^{2} + 1 : - \frac{1271}{9}

(\frac{4}{3}, - \frac{1271}{9})

(- \frac{4}{3}, - \frac{1271}{9}), (\frac{4}{3}, - \frac{1271}{9})

(x + 4)^{3} - 2

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

(0, - 2), (- 4, 4)

f (x) = x^{4} - 128 x^{2} + 4096

f (x) = x^{2 x}

What is the inflection f(x)=x^4-32x^2+1 ?
The inflection f(x)=x^4-32x^2+1 is (-4/(sqrt(3)),-1271/9),(4/(sqrt(3)),-1271/9)