Find the Symmetry f(x)=2/(2x^3-x)

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

Step 1

Determine if the function is odd, even, or neither in order to find the symmetry.

1. If odd, the function is symmetric about the origin.

2. If even, the function is symmetric about the y-axis.

Step 2

Factor $x$ out of $2 x^{3} - x$ .

Factor $x$ out of $2 x^{3}$ .

$f (x) = \frac{2}{x (2 x^{2}) - x}$

Factor $x$ out of $- x$ .

$f (x) = \frac{2}{x (2 x^{2}) + x \cdot - 1}$

Factor $x$ out of $x (2 x^{2}) + x \cdot - 1$ .

$f (x) = \frac{2}{x (2 x^{2} - 1)}$

Step 3

Find $f (- x)$ .

Find $f (- x)$ by substituting $- x$ for all occurrence of $x$ in $f (x)$ .

$f (- x) = \frac{2}{(- x) (2 {(- x)}^{2} - 1)}$

Simplify the denominator.

Apply the product rule to $- x$ .

$f (- x) = \frac{2}{- x (2 ({(- 1)}^{2} x^{2}) - 1)}$

Raise $- 1$ to the power of $2$ .

$f (- x) = \frac{2}{- x (2 (1 x^{2}) - 1)}$

Multiply $x^{2}$ by $1$ .

$f (- x) = \frac{2}{- x (2 x^{2} - 1)}$

Move the negative in front of the fraction.

$f (- x) = - \frac{2}{x (2 x^{2} - 1)}$

Step 4

A function is even if $f (- x) = f (x)$ .

Check if $f (- x) = f (x)$ .

Since $- \frac{2}{x (2 x^{2} - 1)}$ $\neq$ $\frac{2}{x (2 x^{2} - 1)}$ , the function is not even.

The function is not even

Step 5

A function is odd if $f (- x) = - f (x)$ .

Multiply $- 1$ by $\frac{2}{x (2 x^{2} - 1)}$ .

$- f (x) = - \frac{2}{x (2 x^{2} - 1)}$

Since $- \frac{2}{x (2 x^{2} - 1)} = - \frac{2}{x (2 x^{2} - 1)}$ , the function is odd.

The function is odd

Step 6

Since the function is odd, it is symmetric about the origin.

Origin Symmetry

Step 7

Since the function is not even, it is not symmetric about the y-axis.

No y-axis symmetry

Step 8

Determine the symmetry of the function.

Origin symmetry

Step 9

Find the Symmetry f(x)=2/(2x^3-x)

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