Find the Symmetry f(x)=x square root of 2-x^2

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

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

Find $f (- x)$ .

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

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

Apply the product rule to $- x$ .

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

Multiply $- 1$ by ${(- 1)}^{2}$ by adding the exponents.

Move ${(- 1)}^{2}$ .

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

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

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

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $2$ and $1$ .

$f (- x) = - x \sqrt{2 + {(- 1)}^{3} x^{2}}$

Raise $- 1$ to the power of $3$ .

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

Step 3

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

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

Since $- x \sqrt{2 - x^{2}}$ $\neq$ $x \sqrt{2 - x^{2}}$ , the function is not even.

The function is not even

Step 4

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

Remove parentheses.

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

Since $- x \sqrt{2 - x^{2}} = - x \sqrt{2 - x^{2}}$ , the function is odd.

The function is odd

Step 5

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

Origin Symmetry

Step 6

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

No y-axis symmetry

Step 7

Determine the symmetry of the function.

Origin symmetry

Step 8