Calculus Examples

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

Step 1

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

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

Expand the left side.

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Step 3.3.1

Expand $\ln (e^{2 y - 1})$ by moving $2 y - 1$ outside the logarithm.

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

Step 3.3.2

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

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

Step 3.3.3

Multiply $2 y - 1$ by $1$ .

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

Step 3.4

Add $1$ to both sides of the equation.

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

Step 3.5

Divide each term in $2 y = \ln (x) + 1$ by $2$ and simplify.

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Step 3.5.1

Divide each term in $2 y = \ln (x) + 1$ by $2$ .

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

Step 3.5.2

Simplify the left side.

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Step 3.5.2.1

Cancel the common factor of $2$ .

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Step 3.5.2.1.1

Cancel the common factor.

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

Step 3.5.2.1.2

Divide $y$ by $1$ .

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

Step 4

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

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

Step 5

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

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

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

Step 5.2.3

Combine the numerators over the common denominator.

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

Step 5.2.4

Simplify each term.

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Step 5.2.4.1

Use logarithm rules to move $2 x - 1$ out of the exponent.

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

Step 5.2.4.2

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

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

Step 5.2.4.3

Multiply $2 x - 1$ by $1$ .

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

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 $1$ .

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

Step 5.2.5.1.2

Add $2 x$ and $0$ .

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

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.

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

Step 5.2.5.2.2

Divide $x$ by $1$ .

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

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

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

Step 5.3.3

Simplify each term.

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Step 5.3.3.1

Simplify each term.

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Step 5.3.3.1.1

Rewrite $\frac{\ln (x)}{2}$ as $\frac{1}{2} \ln (x)$ .

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

Step 5.3.3.1.2

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

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

Step 5.3.3.2

Apply the distributive property.

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

Step 5.3.3.3

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

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

Step 5.3.3.4

Cancel the common factor of $2$ .

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Step 5.3.3.4.1

Cancel the common factor.

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

Step 5.3.3.4.2

Rewrite the expression.

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

Step 5.3.3.5

Simplify each term.

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Step 5.3.3.5.1

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

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Step 5.3.3.5.1.1

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

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

Step 5.3.3.5.1.2

Cancel the common factor of $2$ .

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Step 5.3.3.5.1.2.1

Cancel the common factor.

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

Step 5.3.3.5.1.2.2

Rewrite the expression.

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

Step 5.3.3.5.2

Simplify.

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

Step 5.3.4

Combine the opposite terms in $\ln (x) + 1 - 1$ .

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Step 5.3.4.1

Subtract $1$ from $1$ .

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

Step 5.3.4.2

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

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

Step 5.3.5

Exponentiation and log are inverse functions.

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

Step 5.4

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

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