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

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

Step 1

Set up the composite result function.

$f (g (x))$

Step 2

Evaluate $f (\frac{x + 3}{x + 1})$ by substituting in the value of $g$ into $f$ .

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

Step 3

Multiply the numerator and denominator of the complex fraction by $x + 1$ .

Multiply $\frac{\frac{x + 3}{x + 1} - 1}{\frac{x + 3}{x + 1} - 3}$ by $\frac{x + 1}{x + 1}$ .

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

Combine.

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

Step 4

Apply the distributive property.

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

Step 5

Simplify by cancelling.

Cancel the common factor of $x + 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x + 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 6

Simplify the numerator.

Apply the distributive property.

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

Move $- 1$ to the left of $x$ .

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

Multiply $- 1$ by $1$ .

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

Rewrite $- 1 x$ as $- x$ .

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

Subtract $x$ from $x$ .

$f (\frac{x + 3}{x + 1}) = \frac{0 + 3 - 1}{x + 3 + (x + 1) \cdot - 3}$

Add $0$ and $3$ .

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

Subtract $1$ from $3$ .

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

Step 7

Simplify the denominator.

Apply the distributive property.

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

Move $- 3$ to the left of $x$ .

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

Multiply $- 3$ by $1$ .

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

Subtract $3 x$ from $x$ .

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

Subtract $3$ from $3$ .

$f (\frac{x + 3}{x + 1}) = \frac{2}{- 2 x + 0}$

Add $- 2 x$ and $0$ .

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

Step 8

Reduce the expression by cancelling the common factors.

Cancel the common factor of $2$ and $- 2$ .

Factor $2$ out of $2$ .

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

Cancel the common factors.

Factor $2$ out of $- 2 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $1$ and $- 1$ .

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

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

Move the negative in front of the fraction.

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

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

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