Find the Product f(x) = square root of 7-x , g(x) = square root of x-7

$f (x) = \sqrt{7 - x}$ , $g (x) = \sqrt{x - 7}$

Step 1

Replace the function designators with the actual functions in $f (x) \cdot (g (x))$ .

$(\sqrt{7 - x}) \cdot (\sqrt{x - 7})$

Step 2

Simplify.

Multiply $(\sqrt{7 - x}) (\sqrt{x - 7})$ .

Combine using the product rule for radicals.

$\sqrt{(7 - x) (x - 7)}$

Rewrite $7$ as $- 1 (- 7)$ .

$\sqrt{(- 1 (- 7) - x) (x - 7)}$

Factor $- 1$ out of $- x$ .

$\sqrt{(- 1 (- 7) - (x)) (x - 7)}$

Factor $- 1$ out of $- 1 (- 7) - (x)$ .

$\sqrt{- 1 (- 7 + x) (x - 7)}$

Reorder terms.

$\sqrt{- 1 (x - 7) (x - 7)}$

Raise $x - 7$ to the power of $1$ .

$\sqrt{- 1 ({(x - 7)}^{1} (x - 7))}$

Raise $x - 7$ to the power of $1$ .

$\sqrt{- 1 ({(x - 7)}^{1} {(x - 7)}^{1})}$

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

$\sqrt{- 1 {(x - 7)}^{1 + 1}}$

Add $1$ and $1$ .

$\sqrt{- 1 {(x - 7)}^{2}}$

Reorder $- 1$ and ${(x - 7)}^{2}$ .

$\sqrt{{(x - 7)}^{2} \cdot - 1}$

Pull terms out from under the radical.

$(x - 7) \sqrt{- 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$(x - 7) i$

Apply the distributive property.

$x i - 7 i$