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

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

Step 1

Set up the composite result function.

$f (g (x))$

Step 2

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

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

Step 3

Multiply the numerator by the reciprocal of the denominator.

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

Step 4

Simplify the denominator.

Write $1$ as a fraction with a common denominator.

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

Combine the numerators over the common denominator.

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

Reorder terms.

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

Rewrite $\frac{x - x + 1}{- x + 1}$ in a factored form.

Subtract $x$ from $x$ .

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

Add $0$ and $1$ .

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

Step 5

Multiply the numerator by the reciprocal of the denominator.

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

Step 6

Multiply $- x + 1$ by $1$ .

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

Step 7

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

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

Step 8

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

Reorder terms.

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

Cancel the common factor.

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

Divide $x$ by $1$ .

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