Find f(g(x)) f(x)=(2-x)/(x+1) g(x)=18-3x

$f (x) = \frac{2 - x}{x + 1}$ $g (x) = 18 - 3 x$

Step 1

Set up the composite result function.

$f (g (x))$

Step 2

Evaluate $f (18 - 3 x)$ by substituting in the value of $g$ into $f$ .

$f (18 - 3 x) = \frac{2 - (18 - 3 x)}{(18 - 3 x) + 1}$

Step 3

Simplify the numerator.

Apply the distributive property.

$f (18 - 3 x) = \frac{2 - 1 \cdot 18 - (- 3 x)}{18 - 3 x + 1}$

Multiply $- 1$ by $18$ .

$f (18 - 3 x) = \frac{2 - 18 - (- 3 x)}{18 - 3 x + 1}$

Multiply $- 3$ by $- 1$ .

$f (18 - 3 x) = \frac{2 - 18 + 3 x}{18 - 3 x + 1}$

Subtract $18$ from $2$ .

$f (18 - 3 x) = \frac{- 16 + 3 x}{18 - 3 x + 1}$

Step 4

Simplify terms.

Add $18$ and $1$ .

$f (18 - 3 x) = \frac{- 16 + 3 x}{- 3 x + 19}$

Rewrite $- 16$ as $- 1 (16)$ .

$f (18 - 3 x) = \frac{- 1 \cdot 16 + 3 x}{- 3 x + 19}$

Factor $- 1$ out of $3 x$ .

$f (18 - 3 x) = \frac{- 1 \cdot 16 - (- 3 x)}{- 3 x + 19}$

Factor $- 1$ out of $- 1 (16) - (- 3 x)$ .

$f (18 - 3 x) = \frac{- 1 (16 - 3 x)}{- 3 x + 19}$

Factor $- 1$ out of $- 3 x$ .

$f (18 - 3 x) = \frac{- 1 (16 - 3 x)}{- (3 x) + 19}$

Rewrite $19$ as $- 1 (- 19)$ .

$f (18 - 3 x) = \frac{- 1 (16 - 3 x)}{- (3 x) - 1 \cdot - 19}$

Factor $- 1$ out of $- (3 x) - 1 (- 19)$ .

$f (18 - 3 x) = \frac{- 1 (16 - 3 x)}{- (3 x - 19)}$

Rewrite $- (3 x - 19)$ as $- 1 (3 x - 19)$ .

$f (18 - 3 x) = \frac{- 1 (16 - 3 x)}{- 1 (3 x - 19)}$

Cancel the common factor.

$f (18 - 3 x) = \frac{- 1 (16 - 3 x)}{- 1 (3 x - 19)}$

Rewrite the expression.

$f (18 - 3 x) = \frac{16 - 3 x}{3 x - 19}$