Calculus Examples

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

Step 1

Write $f (x) = 2 \ln (\frac{x}{2} + 1)$ as an equation.

$y = 2 \ln (\frac{x}{2} + 1)$

Step 2

Interchange the variables.

$x = 2 \ln (\frac{y}{2} + 1)$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $2 \ln (\frac{y}{2} + 1) = x$ .

$2 \ln (\frac{y}{2} + 1) = x$

Step 3.2

Divide each term in $2 \ln (\frac{y}{2} + 1) = x$ by $2$ and simplify.

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Step 3.2.1

Divide each term in $2 \ln (\frac{y}{2} + 1) = x$ by $2$ .

$\frac{2 \ln (\frac{y}{2} + 1)}{2} = \frac{x}{2}$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $2$ .

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Step 3.2.2.1.1

Cancel the common factor.

$\frac{2 \ln (\frac{y}{2} + 1)}{2} = \frac{x}{2}$

Step 3.2.2.1.2

Divide $\ln (\frac{y}{2} + 1)$ by $1$ .

$\ln (\frac{y}{2} + 1) = \frac{x}{2}$

Step 3.3

To solve for $y$ , rewrite the equation using properties of logarithms.

$e^{\ln (\frac{y}{2} + 1)} = e^{\frac{x}{2}}$

Step 3.4

Rewrite $\ln (\frac{y}{2} + 1) = \frac{x}{2}$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{\frac{x}{2}} = \frac{y}{2} + 1$

Step 3.5

Solve for $y$ .

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Step 3.5.1

Rewrite the equation as $\frac{y}{2} + 1 = e^{\frac{x}{2}}$ .

$\frac{y}{2} + 1 = e^{\frac{x}{2}}$

Step 3.5.2

Subtract $1$ from both sides of the equation.

$\frac{y}{2} = e^{\frac{x}{2}} - 1$

Step 3.5.3

Multiply both sides of the equation by $2$ .

$2 \frac{y}{2} = 2 (e^{\frac{x}{2}} - 1)$

Step 3.5.4

Simplify both sides of the equation.

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Step 3.5.4.1

Simplify the left side.

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Step 3.5.4.1.1

Cancel the common factor of $2$ .

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Step 3.5.4.1.1.1

Cancel the common factor.

$2 \frac{y}{2} = 2 (e^{\frac{x}{2}} - 1)$

Step 3.5.4.1.1.2

Rewrite the expression.

$y = 2 (e^{\frac{x}{2}} - 1)$

Step 3.5.4.2

Simplify the right side.

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Step 3.5.4.2.1

Simplify $2 (e^{\frac{x}{2}} - 1)$ .

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Step 3.5.4.2.1.1

Apply the distributive property.

$y = 2 e^{\frac{x}{2}} + 2 \cdot - 1$

Step 3.5.4.2.1.2

Multiply $2$ by $- 1$ .

$y = 2 e^{\frac{x}{2}} - 2$

Step 4

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

$f^{- 1} (x) = 2 e^{\frac{x}{2}} - 2$

Step 5

Verify if $f^{- 1} (x) = 2 e^{\frac{x}{2}} - 2$ is the inverse of $f (x) = 2 \ln (\frac{x}{2} + 1)$ .

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Step 5.1

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

Step 5.2

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

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Step 5.2.1

Set up the composite result function.

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

Step 5.2.2

Evaluate $f^{- 1} (2 \ln (\frac{x}{2} + 1))$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (2 \ln (\frac{x}{2} + 1)) = 2 e^{\frac{2 \ln (\frac{x}{2} + 1)}{2}} - 2$

Step 5.2.3

Simplify each term.

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Step 5.2.3.1

Cancel the common factor of $2$ .

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Step 5.2.3.1.1

Cancel the common factor.

$f^{- 1} (2 \ln (\frac{x}{2} + 1)) = 2 e^{\frac{2 \ln (\frac{x}{2} + 1)}{2}} - 2$

Step 5.2.3.1.2

Divide $\ln (\frac{x}{2} + 1)$ by $1$ .

$f^{- 1} (2 \ln (\frac{x}{2} + 1)) = 2 e^{\ln (\frac{x}{2} + 1)} - 2$

Step 5.2.3.2

Exponentiation and log are inverse functions.

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

Step 5.2.3.3

Apply the distributive property.

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

Step 5.2.3.4

Cancel the common factor of $2$ .

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Step 5.2.3.4.1

Cancel the common factor.

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

Step 5.2.3.4.2

Rewrite the expression.

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

Step 5.2.3.5

Multiply $2$ by $1$ .

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

Step 5.2.4

Combine the opposite terms in $x + 2 - 2$ .

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Step 5.2.4.1

Subtract $2$ from $2$ .

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

Step 5.2.4.2

Add $x$ and $0$ .

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

Step 5.3

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

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Step 5.3.1

Set up the composite result function.

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

Step 5.3.2

Evaluate $f (2 e^{\frac{x}{2}} - 2)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (2 e^{\frac{x}{2}} - 2) = 2 \ln (\frac{2 e^{\frac{x}{2}} - 2}{2} + 1)$

Step 5.3.3

Cancel the common factor of $2 e^{\frac{x}{2}} - 2$ and $2$ .

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Step 5.3.3.1

Factor $2$ out of $2 e^{\frac{x}{2}}$ .

$f (2 e^{\frac{x}{2}} - 2) = 2 \ln (\frac{2 (e^{\frac{x}{2}}) - 2}{2} + 1)$

Step 5.3.3.2

Factor $2$ out of $- 2$ .

$f (2 e^{\frac{x}{2}} - 2) = 2 \ln (\frac{2 (e^{\frac{x}{2}}) + 2 \cdot - 1}{2} + 1)$

Step 5.3.3.3

Factor $2$ out of $2 (e^{\frac{x}{2}}) + 2 (- 1)$ .

$f (2 e^{\frac{x}{2}} - 2) = 2 \ln (\frac{2 (e^{\frac{x}{2}} - 1)}{2} + 1)$

Step 5.3.3.4

Cancel the common factors.

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Step 5.3.3.4.1

Factor $2$ out of $2$ .

$f (2 e^{\frac{x}{2}} - 2) = 2 \ln (\frac{2 (e^{\frac{x}{2}} - 1)}{2 (1)} + 1)$

Step 5.3.3.4.2

Cancel the common factor.

$f (2 e^{\frac{x}{2}} - 2) = 2 \ln (\frac{2 (e^{\frac{x}{2}} - 1)}{2 \cdot 1} + 1)$

Step 5.3.3.4.3

Rewrite the expression.

$f (2 e^{\frac{x}{2}} - 2) = 2 \ln (\frac{e^{\frac{x}{2}} - 1}{1} + 1)$

Step 5.3.3.4.4

Divide $e^{\frac{x}{2}} - 1$ by $1$ .

$f (2 e^{\frac{x}{2}} - 2) = 2 \ln (e^{\frac{x}{2}} - 1 + 1)$

Step 5.3.4

Combine the opposite terms in $e^{\frac{x}{2}} - 1 + 1$ .

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Step 5.3.4.1

Add $- 1$ and $1$ .

$f (2 e^{\frac{x}{2}} - 2) = 2 \ln (e^{\frac{x}{2}} + 0)$

Step 5.3.4.2

Add $e^{\frac{x}{2}}$ and $0$ .

$f (2 e^{\frac{x}{2}} - 2) = 2 \ln (e^{\frac{x}{2}})$

Step 5.3.5

Use logarithm rules to move $\frac{x}{2}$ out of the exponent.

$f (2 e^{\frac{x}{2}} - 2) = 2 (\frac{x}{2} \cdot \ln (e))$

Step 5.3.6

The natural logarithm of $e$ is $1$ .

$f (2 e^{\frac{x}{2}} - 2) = 2 (\frac{x}{2} \cdot 1)$

Step 5.3.7

Multiply $\frac{x}{2}$ by $1$ .

$f (2 e^{\frac{x}{2}} - 2) = 2 (\frac{x}{2})$

Step 5.3.8

Cancel the common factor of $2$ .

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Step 5.3.8.1

Cancel the common factor.

$f (2 e^{\frac{x}{2}} - 2) = 2 (\frac{x}{2})$

Step 5.3.8.2

Rewrite the expression.

$f (2 e^{\frac{x}{2}} - 2) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = 2 e^{\frac{x}{2}} - 2$ is the inverse of $f (x) = 2 \ln (\frac{x}{2} + 1)$ .

$f^{- 1} (x) = 2 e^{\frac{x}{2}} - 2$