Precalculus Examples

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

Step 1

Simplify.

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Step 1.1

Rewrite $1$ as $1^{2}$ .

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

Step 1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = x$ .

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

Step 2

Find $f (- x)$ .

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Step 2.1

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

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

Step 2.2

Remove parentheses.

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

Step 2.3

Apply the product rule to $- x$ .

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

Step 2.4

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

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

Step 2.5

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

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

Step 2.6

Multiply $- - x$ .

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Step 2.6.1

Multiply $- 1$ by $- 1$ .

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

Step 2.6.2

Multiply $x$ by $1$ .

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

Step 3

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

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Step 3.1

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

Step 3.2

Since $x^{2} \sqrt{(1 - x) (1 + x)} = x^{2} \sqrt{(1 + x) (1 - x)}$ , the function is even.

The function is even

Step 4