Precalculus Examples

$f (x) = e^{x - 5} - 5$

Step 1

Write $f (x) = e^{x - 5} - 5$ as an equation.

$y = e^{x - 5} - 5$

Step 2

Interchange the variables.

$x = e^{y - 5} - 5$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $e^{y - 5} - 5 = x$ .

$e^{y - 5} - 5 = x$

Step 3.2

Add $5$ to both sides of the equation.

$e^{y - 5} = x + 5$

Step 3.3

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

$\ln (e^{y - 5}) = \ln (x + 5)$

Step 3.4

Expand the left side.

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Step 3.4.1

Expand $\ln (e^{y - 5})$ by moving $y - 5$ outside the logarithm.

$(y - 5) \ln (e) = \ln (x + 5)$

Step 3.4.2

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

$(y - 5) \cdot 1 = \ln (x + 5)$

Step 3.4.3

Multiply $y - 5$ by $1$ .

$y - 5 = \ln (x + 5)$

Step 3.5

Add $5$ to both sides of the equation.

$y = \ln (x + 5) + 5$

Step 4

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

$f^{- 1} (x) = \ln (x + 5) + 5$

Step 5

Verify if $f^{- 1} (x) = \ln (x + 5) + 5$ is the inverse of $f (x) = e^{x - 5} - 5$ .

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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 $f^{- 1} (e^{x - 5} - 5)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (e^{x - 5} - 5) = \ln ((e^{x - 5} - 5) + 5) + 5$

Step 5.2.3

Combine the opposite terms in $\ln ((e^{x - 5} - 5) + 5) + 5$ .

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Step 5.2.3.1

Add $- 5$ and $5$ .

$f^{- 1} (e^{x - 5} - 5) = \ln (e^{x - 5} + 0) + 5$

Step 5.2.3.2

Add $e^{x - 5}$ and $0$ .

$f^{- 1} (e^{x - 5} - 5) = \ln (e^{x - 5}) + 5$

Step 5.2.4

Simplify each term.

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Step 5.2.4.1

Use logarithm rules to move $x - 5$ out of the exponent.

$f^{- 1} (e^{x - 5} - 5) = (x - 5) \ln (e) + 5$

Step 5.2.4.2

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

$f^{- 1} (e^{x - 5} - 5) = (x - 5) \cdot 1 + 5$

Step 5.2.4.3

Multiply $x - 5$ by $1$ .

$f^{- 1} (e^{x - 5} - 5) = x - 5 + 5$

Step 5.2.5

Combine the opposite terms in $x - 5 + 5$ .

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Step 5.2.5.1

Add $- 5$ and $5$ .

$f^{- 1} (e^{x - 5} - 5) = x + 0$

Step 5.2.5.2

Add $x$ and $0$ .

$f^{- 1} (e^{x - 5} - 5) = x$

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 $f (\ln (x + 5) + 5)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\ln (x + 5) + 5) = e^{(\ln (x + 5) + 5) - 5} - 5$

Step 5.3.3

Combine the opposite terms in $e^{(\ln (x + 5) + 5) - 5} - 5$ .

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Step 5.3.3.1

Subtract $5$ from $5$ .

$f (\ln (x + 5) + 5) = e^{\ln (x + 5) + 0} - 5$

Step 5.3.3.2

Add $\ln (x + 5)$ and $0$ .

$f (\ln (x + 5) + 5) = e^{\ln (x + 5)} - 5$

Step 5.3.4

Exponentiation and log are inverse functions.

$f (\ln (x + 5) + 5) = x + 5 - 5$

Step 5.3.5

Combine the opposite terms in $x + 5 - 5$ .

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Step 5.3.5.1

Subtract $5$ from $5$ .

$f (\ln (x + 5) + 5) = x + 0$

Step 5.3.5.2

Add $x$ and $0$ .

$f (\ln (x + 5) + 5) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \ln (x + 5) + 5$ is the inverse of $f (x) = e^{x - 5} - 5$ .

$f^{- 1} (x) = \ln (x + 5) + 5$