Finite Math Examples

$f (x) = \frac{1}{2} \cdot (5 - e^{x})$

Step 1

Write $f (x) = \frac{1}{2} \cdot (5 - e^{x})$ as an equation.

$y = \frac{1}{2} \cdot (5 - e^{x})$

Step 2

Interchange the variables.

$x = \frac{1}{2} \cdot (5 - e^{y})$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\frac{1}{2} \cdot (5 - e^{y}) = x$ .

$\frac{1}{2} \cdot (5 - e^{y}) = x$

Step 3.2

Multiply both sides of the equation by $2$ .

$2 (\frac{1}{2} \cdot (5 - e^{y})) = 2 x$

Step 3.3

Simplify the left side.

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Step 3.3.1

Simplify $2 (\frac{1}{2} \cdot (5 - e^{y}))$ .

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Step 3.3.1.1

Apply the distributive property.

$2 (\frac{1}{2} \cdot 5 + \frac{1}{2} (- e^{y})) = 2 x$

Step 3.3.1.2

Combine $\frac{1}{2}$ and $5$ .

$2 (\frac{5}{2} + \frac{1}{2} (- e^{y})) = 2 x$

Step 3.3.1.3

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

$2 (\frac{5}{2} - \frac{e^{y}}{2}) = 2 x$

Step 3.3.1.4

Apply the distributive property.

$2 (\frac{5}{2}) + 2 (- \frac{e^{y}}{2}) = 2 x$

Step 3.3.1.5

Cancel the common factor of $2$ .

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Step 3.3.1.5.1

Cancel the common factor.

$2 (\frac{5}{2}) + 2 (- \frac{e^{y}}{2}) = 2 x$

Step 3.3.1.5.2

Rewrite the expression.

$5 + 2 (- \frac{e^{y}}{2}) = 2 x$

Step 3.3.1.6

Cancel the common factor of $2$ .

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Step 3.3.1.6.1

Move the leading negative in $- \frac{e^{y}}{2}$ into the numerator.

$5 + 2 \frac{- e^{y}}{2} = 2 x$

Step 3.3.1.6.2

Cancel the common factor.

$5 + 2 \frac{- e^{y}}{2} = 2 x$

Step 3.3.1.6.3

Rewrite the expression.

$5 - e^{y} = 2 x$

Step 3.4

Subtract $5$ from both sides of the equation.

$- e^{y} = 2 x - 5$

Step 3.5

Divide each term in $- e^{y} = 2 x - 5$ by $- 1$ and simplify.

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Step 3.5.1

Divide each term in $- e^{y} = 2 x - 5$ by $- 1$ .

$\frac{- e^{y}}{- 1} = \frac{2 x}{- 1} + \frac{- 5}{- 1}$

Step 3.5.2

Simplify the left side.

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Step 3.5.2.1

Dividing two negative values results in a positive value.

$\frac{e^{y}}{1} = \frac{2 x}{- 1} + \frac{- 5}{- 1}$

Step 3.5.2.2

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

$e^{y} = \frac{2 x}{- 1} + \frac{- 5}{- 1}$

Step 3.5.3

Simplify the right side.

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Step 3.5.3.1

Simplify each term.

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Step 3.5.3.1.1

Move the negative one from the denominator of $\frac{2 x}{- 1}$ .

$e^{y} = - 1 \cdot (2 x) + \frac{- 5}{- 1}$

Step 3.5.3.1.2

Rewrite $- 1 \cdot (2 x)$ as $- (2 x)$ .

$e^{y} = - (2 x) + \frac{- 5}{- 1}$

Step 3.5.3.1.3

Multiply $2$ by $- 1$ .

$e^{y} = - 2 x + \frac{- 5}{- 1}$

Step 3.5.3.1.4

Divide $- 5$ by $- 1$ .

$e^{y} = - 2 x + 5$

Step 3.6

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

$\ln (e^{y}) = \ln (- 2 x + 5)$

Step 3.7

Expand the left side.

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Step 3.7.1

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

$y \ln (e) = \ln (- 2 x + 5)$

Step 3.7.2

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

$y \cdot 1 = \ln (- 2 x + 5)$

Step 3.7.3

Multiply $y$ by $1$ .

$y = \ln (- 2 x + 5)$

Step 4

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

$f^{- 1} (x) = \ln (- 2 x + 5)$

Step 5

Verify if $f^{- 1} (x) = \ln (- 2 x + 5)$ is the inverse of $f (x) = \frac{1}{2} \cdot (5 - e^{x})$ .

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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 $f^{- 1} (\frac{1}{2} \cdot (5 - e^{x}))$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{1}{2} \cdot (5 - e^{x})) = \ln (- 2 (\frac{1}{2} \cdot (5 - e^{x})) + 5)$

Step 5.2.3

Simplify each term.

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Step 5.2.3.1

Apply the distributive property.

$f^{- 1} (\frac{1}{2} \cdot (5 - e^{x})) = \ln (- 2 (\frac{1}{2} \cdot 5 + \frac{1}{2} \cdot (- e^{x})) + 5)$

Step 5.2.3.2

Combine $\frac{1}{2}$ and $5$ .

$f^{- 1} (\frac{1}{2} \cdot (5 - e^{x})) = \ln (- 2 (\frac{5}{2} + \frac{1}{2} \cdot (- e^{x})) + 5)$

Step 5.2.3.3

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

$f^{- 1} (\frac{1}{2} \cdot (5 - e^{x})) = \ln (- 2 (\frac{5}{2} - \frac{e^{x}}{2}) + 5)$

Step 5.2.3.4

Apply the distributive property.

$f^{- 1} (\frac{1}{2} \cdot (5 - e^{x})) = \ln (- 2 (\frac{5}{2}) - 2 (- \frac{e^{x}}{2}) + 5)$

Step 5.2.3.5

Cancel the common factor of $2$ .

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Step 5.2.3.5.1

Factor $2$ out of $- 2$ .

$f^{- 1} (\frac{1}{2} \cdot (5 - e^{x})) = \ln (2 (- 1) (\frac{5}{2}) - 2 (- \frac{e^{x}}{2}) + 5)$

Step 5.2.3.5.2

Cancel the common factor.

$f^{- 1} (\frac{1}{2} \cdot (5 - e^{x})) = \ln (2 \cdot (- 1 (\frac{5}{2})) - 2 (- \frac{e^{x}}{2}) + 5)$

Step 5.2.3.5.3

Rewrite the expression.

$f^{- 1} (\frac{1}{2} \cdot (5 - e^{x})) = \ln (- 1 \cdot 5 - 2 (- \frac{e^{x}}{2}) + 5)$

Step 5.2.3.6

Multiply $- 1$ by $5$ .

$f^{- 1} (\frac{1}{2} \cdot (5 - e^{x})) = \ln (- 5 - 2 (- \frac{e^{x}}{2}) + 5)$

Step 5.2.3.7

Cancel the common factor of $2$ .

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Step 5.2.3.7.1

Move the leading negative in $- \frac{e^{x}}{2}$ into the numerator.

$f^{- 1} (\frac{1}{2} \cdot (5 - e^{x})) = \ln (- 5 - 2 \frac{- e^{x}}{2} + 5)$

Step 5.2.3.7.2

Factor $2$ out of $- 2$ .

$f^{- 1} (\frac{1}{2} \cdot (5 - e^{x})) = \ln (- 5 + 2 (- 1) (\frac{- e^{x}}{2}) + 5)$

Step 5.2.3.7.3

Cancel the common factor.

$f^{- 1} (\frac{1}{2} \cdot (5 - e^{x})) = \ln (- 5 + 2 \cdot (- 1 \frac{- e^{x}}{2}) + 5)$

Step 5.2.3.7.4

Rewrite the expression.

$f^{- 1} (\frac{1}{2} \cdot (5 - e^{x})) = \ln (- 5 - 1 (- e^{x}) + 5)$

Step 5.2.3.8

Multiply $- 1$ by $- 1$ .

$f^{- 1} (\frac{1}{2} \cdot (5 - e^{x})) = \ln (- 5 + 1 e^{x} + 5)$

Step 5.2.3.9

Multiply $e^{x}$ by $1$ .

$f^{- 1} (\frac{1}{2} \cdot (5 - e^{x})) = \ln (- 5 + e^{x} + 5)$

Step 5.2.4

Combine the opposite terms in $- 5 + e^{x} + 5$ .

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Step 5.2.4.1

Add $- 5$ and $5$ .

$f^{- 1} (\frac{1}{2} \cdot (5 - e^{x})) = \ln (0 + e^{x})$

Step 5.2.4.2

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

$f^{- 1} (\frac{1}{2} \cdot (5 - e^{x})) = \ln (e^{x})$

Step 5.2.5

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

$f^{- 1} (\frac{1}{2} \cdot (5 - e^{x})) = x \ln (e)$

Step 5.2.6

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

$f^{- 1} (\frac{1}{2} \cdot (5 - e^{x})) = x \cdot 1$

Step 5.2.7

Multiply $x$ by $1$ .

$f^{- 1} (\frac{1}{2} \cdot (5 - e^{x})) = x$

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 $f (\ln (- 2 x + 5))$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\ln (- 2 x + 5)) = \frac{1}{2} \cdot (5 - e^{\ln (- 2 x + 5)})$

Step 5.3.3

Simplify each term.

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Step 5.3.3.1

Exponentiation and log are inverse functions.

$f (\ln (- 2 x + 5)) = \frac{1}{2} \cdot (5 - (- 2 x + 5))$

Step 5.3.3.2

Apply the distributive property.

$f (\ln (- 2 x + 5)) = \frac{1}{2} \cdot (5 - (- 2 x) - 1 \cdot 5)$

Step 5.3.3.3

Multiply $- 2$ by $- 1$ .

$f (\ln (- 2 x + 5)) = \frac{1}{2} \cdot (5 + 2 x - 1 \cdot 5)$

Step 5.3.3.4

Multiply $- 1$ by $5$ .

$f (\ln (- 2 x + 5)) = \frac{1}{2} \cdot (5 + 2 x - 5)$

Step 5.3.4

Simplify terms.

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Step 5.3.4.1

Combine the opposite terms in $5 + 2 x - 5$ .

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Step 5.3.4.1.1

Subtract $5$ from $5$ .

$f (\ln (- 2 x + 5)) = \frac{1}{2} \cdot (2 x + 0)$

Step 5.3.4.1.2

Add $2 x$ and $0$ .

$f (\ln (- 2 x + 5)) = \frac{1}{2} \cdot (2 x)$

Step 5.3.4.2

Cancel the common factor of $2$ .

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Step 5.3.4.2.1

Factor $2$ out of $2 x$ .

$f (\ln (- 2 x + 5)) = \frac{1}{2} \cdot (2 (x))$

Step 5.3.4.2.2

Cancel the common factor.

$f (\ln (- 2 x + 5)) = \frac{1}{2} \cdot (2 x)$

Step 5.3.4.2.3

Rewrite the expression.

$f (\ln (- 2 x + 5)) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \ln (- 2 x + 5)$ is the inverse of $f (x) = \frac{1}{2} \cdot (5 - e^{x})$ .

$f^{- 1} (x) = \ln (- 2 x + 5)$