Find g(f(x)) f(x)=(1-x)/(4x) , g(x)=1/(1+4x)

$f (x) = \frac{1 - x}{4 x}$ , $g (x) = \frac{1}{1 + 4 x}$

Step 1

Set up the composite result function.

$g (f (x))$

Step 2

Evaluate $g (\frac{1 - x}{4 x})$ by substituting in the value of $f$ into $g$ .

$g (\frac{1 - x}{4 x}) = \frac{1}{1 + 4 (\frac{1 - x}{4 x})}$

Step 3

Simplify the denominator.

Cancel the common factor of $4$ .

Factor $4$ out of $4 x$ .

$g (\frac{1 - x}{4 x}) = \frac{1}{1 + 4 (\frac{1 - x}{4 (x)})}$

Cancel the common factor.

$g (\frac{1 - x}{4 x}) = \frac{1}{1 + 4 (\frac{1 - x}{4 x})}$

Rewrite the expression.

$g (\frac{1 - x}{4 x}) = \frac{1}{1 + \frac{1 - x}{x}}$

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

$g (\frac{1 - x}{4 x}) = \frac{1}{\frac{x}{x} + \frac{1 - x}{x}}$

Combine the numerators over the common denominator.

$g (\frac{1 - x}{4 x}) = \frac{1}{\frac{x + 1 - x}{x}}$

Reorder terms.

$g (\frac{1 - x}{4 x}) = \frac{1}{\frac{x - x + 1}{x}}$

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

Subtract $x$ from $x$ .

$g (\frac{1 - x}{4 x}) = \frac{1}{\frac{0 + 1}{x}}$

Add $0$ and $1$ .

$g (\frac{1 - x}{4 x}) = \frac{1}{\frac{1}{x}}$

Step 4

Multiply the numerator by the reciprocal of the denominator.

$g (\frac{1 - x}{4 x}) = 1 x$

Step 5

Multiply $x$ by $1$ .

$g (\frac{1 - x}{4 x}) = x$