What is the inflection f(x)=(x^2-1)^3 ?

The inflection f(x)=(x^2-1)^3 is (-1,0),(-1/(sqrt(5)),-64/125),(1/(sqrt(5)),-64/125),(1,0)

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inflection f(x)=(x^2-1)^3

inflection points

f (x) = (x^{2} - 1)^{3}

(- 1, 0), (- \frac{1}{5}, - \frac{64}{125}), (\frac{1}{5}, - \frac{64}{125}), (1, 0)

Solution steps

f^{''} (x) = 6 (5 x^{4} - 6 x^{2} + 1)

Find intervals:

Concave Upward

: - \infty < x < - 1,

Concave Downward

: - 1 < x < - \frac{1}{5},

Concave Upward

: - \frac{1}{5} < x < \frac{1}{5},

Concave Downward

: \frac{1}{5} < x < 1,

Concave Upward

: 1 < x < \infty

Plug

x = - 1

into

(x^{2} - 1)^{3} : 0

(- 1, 0)

Plug

x = - \frac{1}{5}

into

(x^{2} - 1)^{3} : - \frac{64}{125}

(- \frac{1}{5}, - \frac{64}{125})

Plug

x = \frac{1}{5}

into

(x^{2} - 1)^{3} : - \frac{64}{125}

(\frac{1}{5}, - \frac{64}{125})

Plug

x = 1

into

(x^{2} - 1)^{3} : 0

(1, 0)

(- 1, 0), (- \frac{1}{5}, - \frac{64}{125}), (\frac{1}{5}, - \frac{64}{125}), (1, 0)

f (x) = x^{4} + 8 x^{3} + 18 x^{2} - 8

f (x) = \frac{2}{x ^{3}}

h (x) = x^{3} - 5 x^{2} + 4 x - 20

f (x) = \frac{x + 6}{x ( x + 11 )}

f (x) = \frac{x - 8}{x + 5}

What is the inflection f(x)=(x^2-1)^3 ?
The inflection f(x)=(x^2-1)^3 is (-1,0),(-1/(sqrt(5)),-64/125),(1/(sqrt(5)),-64/125),(1,0)