Calculus Examples

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

Step 1

Write $f (x) = \frac{1}{2} \cdot e^{x + 1} - 2$ as an equation.

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

Simplify $\frac{1}{2} \cdot e^{y + 1} - 2$ .

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Step 3.2.1

Combine $\frac{1}{2}$ and $e^{y + 1}$ .

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

Step 3.2.2

To write $- 2$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 3.2.3

Combine $- 2$ and $\frac{2}{2}$ .

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

Step 3.2.4

Combine the numerators over the common denominator.

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

Step 3.2.5

Multiply $- 2$ by $2$ .

$\frac{e^{y + 1} - 4}{2} = x$

Step 3.3

Multiply both sides of the equation by $2$ .

$2 \frac{e^{y + 1} - 4}{2} = 2 x$

Step 3.4

Simplify the left side.

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Step 3.4.1

Cancel the common factor of $2$ .

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Step 3.4.1.1

Cancel the common factor.

$2 \frac{e^{y + 1} - 4}{2} = 2 x$

Step 3.4.1.2

Rewrite the expression.

$e^{y + 1} - 4 = 2 x$

Step 3.5

Add $4$ to both sides of the equation.

$e^{y + 1} = 2 x + 4$

Step 3.6

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

$\ln (e^{y + 1}) = \ln (2 x + 4)$

Step 3.7

Expand the left side.

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Step 3.7.1

Expand $\ln (e^{y + 1})$ by moving $y + 1$ outside the logarithm.

$(y + 1) \ln (e) = \ln (2 x + 4)$

Step 3.7.2

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

$(y + 1) \cdot 1 = \ln (2 x + 4)$

Step 3.7.3

Multiply $y + 1$ by $1$ .

$y + 1 = \ln (2 x + 4)$

Step 3.8

Subtract $1$ from both sides of the equation.

$y = \ln (2 x + 4) - 1$

Step 4

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

$f^{- 1} (x) = \ln (2 x + 4) - 1$

Step 5

Verify if $f^{- 1} (x) = \ln (2 x + 4) - 1$ is the inverse of $f (x) = \frac{1}{2} \cdot e^{x + 1} - 2$ .

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

$f^{- 1} (\frac{1}{2} \cdot e^{x + 1} - 2) = \ln (2 (\frac{1}{2} \cdot e^{x + 1} - 2) + 4) - 1$

Step 5.2.3

Simplify each term.

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Step 5.2.3.1

Simplify each term.

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Step 5.2.3.1.1

Combine $\frac{1}{2}$ and $e^{x + 1}$ .

$f^{- 1} (\frac{1}{2} \cdot e^{x + 1} - 2) = \ln (2 (\frac{e^{x + 1}}{2} - 2) + 4) - 1$

Step 5.2.3.1.2

To write $- 2$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f^{- 1} (\frac{1}{2} \cdot e^{x + 1} - 2) = \ln (2 (\frac{e^{x + 1}}{2} - 2 \cdot \frac{2}{2}) + 4) - 1$

Step 5.2.3.1.3

Combine $- 2$ and $\frac{2}{2}$ .

$f^{- 1} (\frac{1}{2} \cdot e^{x + 1} - 2) = \ln (2 (\frac{e^{x + 1}}{2} + \frac{- 2 \cdot 2}{2}) + 4) - 1$

Step 5.2.3.1.4

Combine the numerators over the common denominator.

$f^{- 1} (\frac{1}{2} \cdot e^{x + 1} - 2) = \ln (2 (\frac{e^{x + 1} - 2 \cdot 2}{2}) + 4) - 1$

Step 5.2.3.1.5

Multiply $- 2$ by $2$ .

$f^{- 1} (\frac{1}{2} \cdot e^{x + 1} - 2) = \ln (2 (\frac{e^{x + 1} - 4}{2}) + 4) - 1$

Step 5.2.3.1.6

Cancel the common factor of $2$ .

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Step 5.2.3.1.6.1

Cancel the common factor.

$f^{- 1} (\frac{1}{2} \cdot e^{x + 1} - 2) = \ln (2 (\frac{e^{x + 1} - 4}{2}) + 4) - 1$

Step 5.2.3.1.6.2

Rewrite the expression.

$f^{- 1} (\frac{1}{2} \cdot e^{x + 1} - 2) = \ln (e^{x + 1} - 4 + 4) - 1$

Step 5.2.3.2

Combine the opposite terms in $e^{x + 1} - 4 + 4$ .

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Step 5.2.3.2.1

Add $- 4$ and $4$ .

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

Step 5.2.3.2.2

Add $e^{x + 1}$ and $0$ .

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

Step 5.2.3.3

Use logarithm rules to move $x + 1$ out of the exponent.

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

Step 5.2.3.4

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

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

Step 5.2.3.5

Multiply $x + 1$ by $1$ .

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

Step 5.2.4

Combine the opposite terms in $x + 1 - 1$ .

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Step 5.2.4.1

Subtract $1$ from $1$ .

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

Step 5.2.4.2

Add $x$ and $0$ .

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

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 $f (\ln (2 x + 4) - 1)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\ln (2 x + 4) - 1) = \frac{1}{2} \cdot e^{(\ln (2 x + 4) - 1) + 1} - 2$

Step 5.3.3

Combine the opposite terms in $\frac{1}{2} \cdot e^{(\ln (2 x + 4) - 1) + 1} - 2$ .

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Step 5.3.3.1

Add $- 1$ and $1$ .

$f (\ln (2 x + 4) - 1) = \frac{1}{2} \cdot e^{\ln (2 x + 4) + 0} - 2$

Step 5.3.3.2

Add $\ln (2 x + 4)$ and $0$ .

$f (\ln (2 x + 4) - 1) = \frac{1}{2} \cdot e^{\ln (2 x + 4)} - 2$

Step 5.3.4

Simplify each term.

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Step 5.3.4.1

Exponentiation and log are inverse functions.

$f (\ln (2 x + 4) - 1) = \frac{1}{2} \cdot (2 x + 4) - 2$

Step 5.3.4.2

Apply the distributive property.

$f (\ln (2 x + 4) - 1) = \frac{1}{2} \cdot (2 x) + \frac{1}{2} \cdot 4 - 2$

Step 5.3.4.3

Cancel the common factor of $2$ .

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Step 5.3.4.3.1

Factor $2$ out of $2 x$ .

$f (\ln (2 x + 4) - 1) = \frac{1}{2} \cdot (2 (x)) + \frac{1}{2} \cdot 4 - 2$

Step 5.3.4.3.2

Cancel the common factor.

$f (\ln (2 x + 4) - 1) = \frac{1}{2} \cdot (2 x) + \frac{1}{2} \cdot 4 - 2$

Step 5.3.4.3.3

Rewrite the expression.

$f (\ln (2 x + 4) - 1) = x + \frac{1}{2} \cdot 4 - 2$

Step 5.3.4.4

Cancel the common factor of $2$ .

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Step 5.3.4.4.1

Factor $2$ out of $4$ .

$f (\ln (2 x + 4) - 1) = x + \frac{1}{2} \cdot (2 (2)) - 2$

Step 5.3.4.4.2

Cancel the common factor.

$f (\ln (2 x + 4) - 1) = x + \frac{1}{2} \cdot (2 \cdot 2) - 2$

Step 5.3.4.4.3

Rewrite the expression.

$f (\ln (2 x + 4) - 1) = x + 2 - 2$

Step 5.3.5

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

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Step 5.3.5.1

Subtract $2$ from $2$ .

$f (\ln (2 x + 4) - 1) = x + 0$

Step 5.3.5.2

Add $x$ and $0$ .

$f (\ln (2 x + 4) - 1) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \ln (2 x + 4) - 1$ is the inverse of $f (x) = \frac{1}{2} \cdot e^{x + 1} - 2$ .

$f^{- 1} (x) = \ln (2 x + 4) - 1$