Calculus Examples

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

Step 1

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

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

Multiply both sides of the equation by $2$ .

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

Step 3.3

Simplify the left side.

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Step 3.3.1

Simplify $2 (\frac{1}{2} \cdot \ln (2 y - 1))$ .

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Step 3.3.1.1

Combine $\frac{1}{2}$ and $\ln (2 y - 1)$ .

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

Step 3.3.1.2

Cancel the common factor of $2$ .

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Step 3.3.1.2.1

Cancel the common factor.

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

Step 3.3.1.2.2

Rewrite the expression.

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

Step 3.4

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

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

Step 3.5

Rewrite $\ln (2 y - 1) = 2 x$ 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 $b^{y} = x$ .

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

Step 3.6

Solve for $y$ .

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Step 3.6.1

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

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

Step 3.6.2

Add $1$ to both sides of the equation.

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

Step 3.6.3

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

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Step 3.6.3.1

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

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

Step 3.6.3.2

Simplify the left side.

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Step 3.6.3.2.1

Cancel the common factor of $2$ .

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Step 3.6.3.2.1.1

Cancel the common factor.

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

Step 3.6.3.2.1.2

Divide $y$ by $1$ .

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

Step 4

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

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

Step 5

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

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

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

Step 5.2.3

Combine the numerators over the common denominator.

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

Step 5.2.4

Simplify each term.

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Step 5.2.4.1

Simplify $\frac{1}{2} (\ln (2 x - 1))$ by moving $\frac{1}{2}$ inside the logarithm.

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

Step 5.2.4.2

Simplify $2 \ln ({(2 x - 1)}^{\frac{1}{2}})$ by moving $2$ inside the logarithm.

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

Step 5.2.4.3

Exponentiation and log are inverse functions.

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

Step 5.2.4.4

Multiply the exponents in ${({(2 x - 1)}^{\frac{1}{2}})}^{2}$ .

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Step 5.2.4.4.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 5.2.4.4.2

Cancel the common factor of $2$ .

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Step 5.2.4.4.2.1

Cancel the common factor.

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

Step 5.2.4.4.2.2

Rewrite the expression.

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

Step 5.2.4.5

Simplify.

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

Step 5.2.5

Simplify terms.

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Step 5.2.5.1

Combine the opposite terms in $2 x - 1 + 1$ .

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Step 5.2.5.1.1

Add $- 1$ and $1$ .

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

Step 5.2.5.1.2

Add $2 x$ and $0$ .

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

Step 5.2.5.2

Cancel the common factor of $2$ .

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Step 5.2.5.2.1

Cancel the common factor.

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

Step 5.2.5.2.2

Divide $x$ by $1$ .

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

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

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

Step 5.3.3

Simplify each term.

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Step 5.3.3.1

Apply the distributive property.

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

Step 5.3.3.2

Cancel the common factor of $2$ .

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Step 5.3.3.2.1

Cancel the common factor.

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

Step 5.3.3.2.2

Rewrite the expression.

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

Step 5.3.3.3

Cancel the common factor of $2$ .

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Step 5.3.3.3.1

Cancel the common factor.

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

Step 5.3.3.3.2

Rewrite the expression.

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

Step 5.3.4

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

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Step 5.3.4.1

Subtract $1$ from $1$ .

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

Step 5.3.4.2

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

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

Step 5.3.5

Use logarithm rules to move $2 x$ out of the exponent.

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

Step 5.3.6

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

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

Step 5.3.7

Multiply $2$ by $1$ .

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

Step 5.3.8

Cancel the common factor of $2$ .

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Step 5.3.8.1

Factor $2$ out of $2 x$ .

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

Step 5.3.8.2

Cancel the common factor.

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

Step 5.3.8.3

Rewrite the expression.

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

Step 5.4

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

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