Find g(f(x)) f(x)=x^2+1 , g(x) = square root of 2-x

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

Step 1

Set up the composite result function.

$g (f (x))$

Step 2

Evaluate $g (x^{2} + 1)$ by substituting in the value of $f$ into $g$ .

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

Step 3

Apply the distributive property.

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

Step 4

Simplify the expression.

Multiply $- 1$ by $1$ .

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

Subtract $1$ from $2$ .

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

Step 5

Rewrite $- x^{2} + 1$ in a factored form.

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

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

Reorder $- x^{2}$ and $1^{2}$ .

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

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 $b = x$ .

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