Precalculus Examples

$f (x) = \sqrt{x}$ , $g (x) = 49 - x^{2}$

Step 1

Set up the composite result function.

$f (g (x))$

Step 2

Evaluate $f (49 - x^{2})$ by substituting in the value of $g$ into $f$ .

$f (49 - x^{2}) = \sqrt{49 - x^{2}}$

Step 3

Remove parentheses.

$f (49 - x^{2}) = \sqrt{49 - x^{2}}$

Step 4

Rewrite $49$ as $7^{2}$ .

$f (49 - x^{2}) = \sqrt{7^{2} - x^{2}}$

Step 5

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 7$ and $b = x$ .

$f (49 - x^{2}) = \sqrt{(7 + x) (7 - x)}$