Precalculus Examples

$f (x) = 2 e^{x + 1}$

Step 1

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

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

Divide each term in $2 e^{y + 1} = x$ by $2$ and simplify.

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Step 3.2.1

Divide each term in $2 e^{y + 1} = x$ by $2$ .

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

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.

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

Step 3.2.2.1.2

Divide $e^{y + 1}$ by $1$ .

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

Step 3.3

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

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

Step 3.4

Expand the left side.

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Step 3.4.1

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

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

Step 3.4.2

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

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

Step 3.4.3

Multiply $y + 1$ by $1$ .

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

Step 3.5

Subtract $1$ from both sides of the equation.

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

Step 4

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

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

Step 5

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

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

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

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.

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

Step 5.2.3.1.2

Divide $e^{x + 1}$ by $1$ .

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

Step 5.2.3.2

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

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

Step 5.2.3.3

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

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

Step 5.2.3.4

Multiply $x + 1$ by $1$ .

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

Step 5.2.4

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

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Step 5.2.4.1

Subtract $1$ from $1$ .

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

Step 5.2.4.2

Add $x$ and $0$ .

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

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

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

Step 5.3.3

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

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Step 5.3.3.1

Add $- 1$ and $1$ .

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

Step 5.3.3.2

Add $\ln (\frac{x}{2})$ and $0$ .

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

Step 5.3.4

Exponentiation and log are inverse functions.

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

Step 5.3.5

Cancel the common factor of $2$ .

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Step 5.3.5.1

Cancel the common factor.

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

Step 5.3.5.2

Rewrite the expression.

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

Step 5.4

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

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