What is the inflection f(x)=0.2x^2+1/(2x) ?

The inflection f(x)=0.2x^2+1/(2x) is (-\sqrt[3]{5/2},-8.0E-17)

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inflection f(x)=0.2x^2+1/(2x)

inflection points

f (x) = 0.2 x^{2} + \frac{1}{2 x}

Solution steps

Find where

f^{''} (x)

is equal to zero or undefined

Identify inflection points not in domain or where

f (x)

is not continuous

Domain of

0.2 x^{2} + \frac{1}{2 x} : x < 0 or x > 0

f (x)

is not defined at

x = 0

, therefore:

Find intervals:

Concave Upward

Concave Downward

Concave Upward

: 0 < x < \infty

Plug

into

0.2 x^{2} + \frac{1}{2 x} : - 8.0 E - 17

(- 6, 1), (1, 8)

f (x) = \frac{4 x + 9}{7 x}

f (x) = 2 x^{5} + x^{7}

- \frac{1}{5} 3^{x} - 2

What is the inflection f(x)=0.2x^2+1/(2x) ?
The inflection f(x)=0.2x^2+1/(2x) is (-\sqrt[3]{5/2},-8.0E-17)