Precalculus Examples

$f (x) = \frac{4^{x}}{1 + 4^{x}}$

Step 1

Write $f (x) = \frac{4^{x}}{1 + 4^{x}}$ as an equation.

$y = \frac{4^{x}}{1 + 4^{x}}$

Step 2

Interchange the variables.

$x = \frac{4^{y}}{1 + 4^{y}}$

Step 3

Solve for $y$ .

Tap for more steps...

Step 3.1

Rewrite the equation as $\frac{4^{y}}{1 + 4^{y}} = x$ .

$\frac{4^{y}}{1 + 4^{y}} = x$

Step 3.2

Multiply both sides by $1 + 4^{y}$ .

$\frac{4^{y}}{1 + 4^{y}} (1 + 4^{y}) = x (1 + 4^{y})$

Step 3.3

Simplify.

Tap for more steps...

Step 3.3.1

Simplify the left side.

Tap for more steps...

Step 3.3.1.1

Cancel the common factor of $1 + 4^{y}$ .

Tap for more steps...

Step 3.3.1.1.1

Cancel the common factor.

$\frac{4^{y}}{1 + 4^{y}} (1 + 4^{y}) = x (1 + 4^{y})$

Step 3.3.1.1.2

Rewrite the expression.

$4^{y} = x (1 + 4^{y})$

Step 3.3.2

Simplify the right side.

Tap for more steps...

Step 3.3.2.1

Simplify $x (1 + 4^{y})$ .

Tap for more steps...

Step 3.3.2.1.1

Apply the distributive property.

$4^{y} = x \cdot 1 + x \cdot 4^{y}$

Step 3.3.2.1.2

Simplify the expression.

Tap for more steps...

Step 3.3.2.1.2.1

Multiply $x$ by $1$ .

$4^{y} = x + x \cdot 4^{y}$

Step 3.3.2.1.2.2

Reorder $x$ and $4^{y}$ .

$4^{y} = x + 4^{y} \cdot x$

Step 3.4

Solve for $y$ .

Tap for more steps...

Step 3.4.1

Reorder factors in $x + 4^{y} x$ .

$4^{y} = x + x \cdot 4^{y}$

Step 3.4.2

Subtract $x \cdot 4^{y}$ from both sides of the equation.

$4^{y} - x \cdot 4^{y} = x$

Step 3.4.3

Factor $4^{y}$ out of $4^{y} - x \cdot 4^{y}$ .

Tap for more steps...

Step 3.4.3.1

Multiply by $1$ .

$4^{y} \cdot 1 - x \cdot 4^{y} = x$

Step 3.4.3.2

Factor $4^{y}$ out of $- x \cdot 4^{y}$ .

$4^{y} \cdot 1 + 4^{y} \cdot (- x) = x$

Step 3.4.3.3

Factor $4^{y}$ out of $4^{y} \cdot 1 + 4^{y} \cdot (- x)$ .

$4^{y} (1 - x) = x$

Step 3.4.4

Divide each term in $4^{y} (1 - x) = x$ by $1 - x$ and simplify.

Tap for more steps...

Step 3.4.4.1

Divide each term in $4^{y} (1 - x) = x$ by $1 - x$ .

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

Step 3.4.4.2

Simplify the left side.

Tap for more steps...

Step 3.4.4.2.1

Cancel the common factor of $1 - x$ .

Tap for more steps...

Step 3.4.4.2.1.1

Cancel the common factor.

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

Step 3.4.4.2.1.2

Divide $4^{y}$ by $1$ .

$4^{y} = \frac{x}{1 - x}$

Step 3.4.5

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (4^{y}) = \ln (\frac{x}{1 - x})$

Step 3.4.6

Expand $\ln (4^{y})$ by moving $y$ outside the logarithm.

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

Step 3.4.7

Divide each term in $y \ln (4) = \ln (\frac{x}{1 - x})$ by $\ln (4)$ and simplify.

Tap for more steps...

Step 3.4.7.1

Divide each term in $y \ln (4) = \ln (\frac{x}{1 - x})$ by $\ln (4)$ .

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

Step 3.4.7.2

Simplify the left side.

Tap for more steps...

Step 3.4.7.2.1

Cancel the common factor of $\ln (4)$ .

Tap for more steps...

Step 3.4.7.2.1.1

Cancel the common factor.

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

Step 3.4.7.2.1.2

Divide $y$ by $1$ .

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

Step 4

Replace $y$ with $f^{- 1} (x)$ to show the final answer.

$f^{- 1} (x) = \frac{\ln (\frac{x}{1 - x})}{\ln (4)}$

Step 5

Verify if $f^{- 1} (x) = \frac{\ln (\frac{x}{1 - x})}{\ln (4)}$ is the inverse of $f (x) = \frac{4^{x}}{1 + 4^{x}}$ .

Tap for more steps...

Step 5.1

To verify the inverse, check if $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Step 5.2

Evaluate $f^{- 1} (f (x))$ .

Tap for more steps...

Step 5.2.1

Set up the composite result function.

$f^{- 1} (f (x))$

Step 5.2.2

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

$f^{- 1} (\frac{4^{x}}{1 + 4^{x}}) = \frac{\ln (\frac{\frac{4^{x}}{1 + 4^{x}}}{1 - (\frac{4^{x}}{1 + 4^{x}})})}{\ln (4)}$

Step 5.2.3

Simplify the denominator.

Tap for more steps...

Step 5.2.3.1

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

$f^{- 1} (\frac{4^{x}}{1 + 4^{x}}) = \frac{\ln (\frac{\frac{4^{x}}{1 + 4^{x}}}{\frac{1 + 4^{x}}{1 + 4^{x}} - \frac{4^{x}}{1 + 4^{x}}})}{\ln (4)}$

Step 5.2.3.2

Combine the numerators over the common denominator.

$f^{- 1} (\frac{4^{x}}{1 + 4^{x}}) = \frac{\ln (\frac{\frac{4^{x}}{1 + 4^{x}}}{\frac{1 + 4^{x} - 4^{x}}{1 + 4^{x}}})}{\ln (4)}$

Step 5.2.3.3

Rewrite $\frac{1 + 4^{x} - 4^{x}}{1 + 4^{x}}$ in a factored form.

Tap for more steps...

Step 5.2.3.3.1

Subtract $4^{x}$ from $4^{x}$ .

$f^{- 1} (\frac{4^{x}}{1 + 4^{x}}) = \frac{\ln (\frac{\frac{4^{x}}{1 + 4^{x}}}{\frac{1 + 0}{1 + 4^{x}}})}{\ln (4)}$

Step 5.2.3.3.2

Add $1$ and $0$ .

$f^{- 1} (\frac{4^{x}}{1 + 4^{x}}) = \frac{\ln (\frac{\frac{4^{x}}{1 + 4^{x}}}{\frac{1}{1 + 4^{x}}})}{\ln (4)}$

Step 5.2.4

Simplify the numerator.

Tap for more steps...

Step 5.2.4.1

Multiply the numerator by the reciprocal of the denominator.

$f^{- 1} (\frac{4^{x}}{1 + 4^{x}}) = \frac{\ln (\frac{4^{x}}{1 + 4^{x}} \cdot (1 + 4^{x}))}{\ln (4)}$

Step 5.2.4.2

Cancel the common factor of $1 + 4^{x}$ .

Tap for more steps...

Step 5.2.4.2.1

Cancel the common factor.

$f^{- 1} (\frac{4^{x}}{1 + 4^{x}}) = \frac{\ln (\frac{4^{x}}{1 + 4^{x}} \cdot (1 + 4^{x}))}{\ln (4)}$

Step 5.2.4.2.2

Rewrite the expression.

$f^{- 1} (\frac{4^{x}}{1 + 4^{x}}) = \frac{\ln (4^{x})}{\ln (4)}$

Step 5.2.5

Expand $\ln (4^{x})$ by moving $x$ outside the logarithm.

$f^{- 1} (\frac{4^{x}}{1 + 4^{x}}) = \frac{x \ln (4)}{\ln (4)}$

Step 5.2.6

Cancel the common factor of $\ln (4)$ .

Tap for more steps...

Step 5.2.6.1

Cancel the common factor.

$f^{- 1} (\frac{4^{x}}{1 + 4^{x}}) = \frac{x \ln (4)}{\ln (4)}$

Step 5.2.6.2

Divide $x$ by $1$ .

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

Step 5.3

Evaluate $f (f^{- 1} (x))$ .

Tap for more steps...

Step 5.3.1

Set up the composite result function.

$f (f^{- 1} (x))$

Step 5.3.2

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

$f (\frac{\ln (\frac{x}{1 - x})}{\ln (4)}) = \frac{4^{\frac{\ln (\frac{x}{1 - x})}{\ln (4)}}}{1 + 4^{\frac{\ln (\frac{x}{1 - x})}{\ln (4)}}}$

Step 5.3.3

Simplify the numerator.

Tap for more steps...

Step 5.3.3.1

Use the change of base rule $\frac{\log_{b} (x)}{\log_{b} (a)} = \log_{a} (x)$ .

$f (\frac{\ln (\frac{x}{1 - x})}{\ln (4)}) = \frac{4^{\log_{4} (\frac{x}{1 - x})}}{1 + 4^{\frac{\ln (\frac{x}{1 - x})}{\ln (4)}}}$

Step 5.3.3.2

Exponentiation and log are inverse functions.

$f (\frac{\ln (\frac{x}{1 - x})}{\ln (4)}) = \frac{\frac{x}{1 - x}}{1 + 4^{\frac{\ln (\frac{x}{1 - x})}{\ln (4)}}}$

Step 5.3.4

Simplify the denominator.

Tap for more steps...

Step 5.3.4.1

Use the change of base rule $\frac{\log_{b} (x)}{\log_{b} (a)} = \log_{a} (x)$ .

$f (\frac{\ln (\frac{x}{1 - x})}{\ln (4)}) = \frac{\frac{x}{1 - x}}{1 + 4^{\log_{4} (\frac{x}{1 - x})}}$

Step 5.3.4.2

Exponentiation and log are inverse functions.

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

Step 5.3.4.3

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

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

Step 5.3.4.4

Combine the numerators over the common denominator.

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

Step 5.3.4.5

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

Tap for more steps...

Step 5.3.4.5.1

Add $- x$ and $x$ .

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

Step 5.3.4.5.2

Add $1$ and $0$ .

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

Step 5.3.5

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{\ln (\frac{x}{1 - x})}{\ln (4)}) = \frac{x}{1 - x} \cdot (1 - x)$

Step 5.3.6

Cancel the common factor of $1 - x$ .

Tap for more steps...

Step 5.3.6.1

Cancel the common factor.

$f (\frac{\ln (\frac{x}{1 - x})}{\ln (4)}) = \frac{x}{1 - x} \cdot (1 - x)$

Step 5.3.6.2

Rewrite the expression.

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

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{\ln (\frac{x}{1 - x})}{\ln (4)}$ is the inverse of $f (x) = \frac{4^{x}}{1 + 4^{x}}$ .

$f^{- 1} (x) = \frac{\ln (\frac{x}{1 - x})}{\ln (4)}$