Calculus Examples

f (x) = 3 e^{2 x} + 1

$f (x) = 3 e^{2 x} + 1$

Step 1

Write

$f (x) = 3 e^{2 x} + 1$ as an equation.

$y = 3 e^{2 x} + 1$

Step 2

Interchange the variables.

$x = 3 e^{2 y} + 1$

Step 3

Solve for

$y$ .

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Step 3.1

Rewrite the equation as

$3 e^{2 y} + 1 = x$ .

$3 e^{2 y} + 1 = x$

Step 3.2

Subtract

$1$ from both sides of the equation.

$3 e^{2 y} = x - 1$

Step 3.3

Divide each term in

$3 e^{2 y} = x - 1$ by

$3$ and simplify.

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Step 3.3.1

Divide each term in

$3 e^{2 y} = x - 1$ by

$3$ .

$\frac{3 e^{2 y}}{3} = \frac{x}{3} + \frac{- 1}{3}$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of

$3$ .

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Step 3.3.2.1.1

Cancel the common factor.

$\frac{3 e^{2 y}}{3} = \frac{x}{3} + \frac{- 1}{3}$

Step 3.3.2.1.2

Divide

$e^{2 y}$ by

$1$ .

$e^{2 y} = \frac{x}{3} + \frac{- 1}{3}$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Move the negative in front of the fraction.

$e^{2 y} = \frac{x}{3} - \frac{1}{3}$

Step 3.4

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

$\ln (e^{2 y}) = \ln (\frac{x}{3} - \frac{1}{3})$

Step 3.5

Expand the left side.

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Step 3.5.1

Expand

$\ln (e^{2 y})$ by moving

$2 y$ outside the logarithm.

$2 y \ln (e) = \ln (\frac{x}{3} - \frac{1}{3})$

Step 3.5.2

The natural logarithm of

$e$ is

$1$ .

$2 y \cdot 1 = \ln (\frac{x}{3} - \frac{1}{3})$

Step 3.5.3

Multiply

$2$ by

$1$ .

$2 y = \ln (\frac{x}{3} - \frac{1}{3})$

Step 3.6

Divide each term in

$2 y = \ln (\frac{x}{3} - \frac{1}{3})$ by

$2$ and simplify.

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Step 3.6.1

Divide each term in

$2 y = \ln (\frac{x}{3} - \frac{1}{3})$ by

$2$ .

$\frac{2 y}{2} = \frac{\ln (\frac{x}{3} - \frac{1}{3})}{2}$

Step 3.6.2

Simplify the left side.

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Step 3.6.2.1

Cancel the common factor of

$2$ .

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Step 3.6.2.1.1

Cancel the common factor.

$\frac{2 y}{2} = \frac{\ln (\frac{x}{3} - \frac{1}{3})}{2}$

Step 3.6.2.1.2

Divide

$y$ by

$1$ .

$y = \frac{\ln (\frac{x}{3} - \frac{1}{3})}{2}$

Step 4

Replace

$y$ with

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

$f^{- 1} (x) = \frac{\ln (\frac{x}{3} - \frac{1}{3})}{2}$

Step 5

Verify if

$f^{- 1} (x) = \frac{\ln (\frac{x}{3} - \frac{1}{3})}{2}$ is the inverse of

$f (x) = 3 e^{2 x} + 1$ .

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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

$f^{- 1} (3 e^{2 x} + 1)$ by substituting in the value of

$f$ into

$f^{- 1}$ .

$f^{- 1} (3 e^{2 x} + 1) = \frac{\ln (\frac{3 e^{2 x} + 1}{3} - \frac{1}{3})}{2}$

Step 5.2.3

Rewrite

$\frac{\ln (\frac{3 e^{2 x} + 1}{3} - \frac{1}{3})}{2}$ as

$\frac{1}{2} \ln (\frac{1}{3} (3 e^{2 x} + 1) - \frac{1}{3})$ .

$f^{- 1} (3 e^{2 x} + 1) = \frac{1}{2} \cdot \ln (\frac{1}{3} \cdot (3 e^{2 x} + 1) - \frac{1}{3})$

Step 5.2.4

Simplify

$\frac{1}{2} \ln (\frac{1}{3} (3 e^{2 x} + 1) - \frac{1}{3})$ by moving

$\frac{1}{2}$ inside the logarithm.

$f^{- 1} (3 e^{2 x} + 1) = \ln ({(\frac{1}{3} \cdot (3 e^{2 x} + 1) - \frac{1}{3})}^{\frac{1}{2}})$

Step 5.2.5

Simplify each term.

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Step 5.2.5.1

Apply the distributive property.

$f^{- 1} (3 e^{2 x} + 1) = \ln ({(\frac{1}{3} \cdot (3 e^{2 x}) + \frac{1}{3} \cdot 1 - \frac{1}{3})}^{\frac{1}{2}})$

Step 5.2.5.2

Cancel the common factor of

$3$ .

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Step 5.2.5.2.1

Factor

$3$ out of

$3 e^{2 x}$ .

$f^{- 1} (3 e^{2 x} + 1) = \ln ({(\frac{1}{3} \cdot (3 (e^{2 x})) + \frac{1}{3} \cdot 1 - \frac{1}{3})}^{\frac{1}{2}})$

Step 5.2.5.2.2

Cancel the common factor.

$f^{- 1} (3 e^{2 x} + 1) = \ln ({(\frac{1}{3} \cdot (3 e^{2 x}) + \frac{1}{3} \cdot 1 - \frac{1}{3})}^{\frac{1}{2}})$

Step 5.2.5.2.3

Rewrite the expression.

$f^{- 1} (3 e^{2 x} + 1) = \ln ({(e^{2 x} + \frac{1}{3} \cdot 1 - \frac{1}{3})}^{\frac{1}{2}})$

Step 5.2.5.3

Multiply

$\frac{1}{3}$ by

$1$ .

$f^{- 1} (3 e^{2 x} + 1) = \ln ({(e^{2 x} + \frac{1}{3} - \frac{1}{3})}^{\frac{1}{2}})$

Step 5.2.6

Simplify by adding terms.

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Step 5.2.6.1

Combine the opposite terms in

$e^{2 x} + \frac{1}{3} - \frac{1}{3}$ .

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Step 5.2.6.1.1

Combine the numerators over the common denominator.

$f^{- 1} (3 e^{2 x} + 1) = \ln ({(e^{2 x} + \frac{1 - 1}{3})}^{\frac{1}{2}})$

Step 5.2.6.1.2

Subtract

$1$ from

$1$ .

$f^{- 1} (3 e^{2 x} + 1) = \ln ({(e^{2 x} + \frac{0}{3})}^{\frac{1}{2}})$

Step 5.2.6.1.3

Divide

$0$ by

$3$ .

$f^{- 1} (3 e^{2 x} + 1) = \ln ({(e^{2 x} + 0)}^{\frac{1}{2}})$

Step 5.2.6.1.4

Add

$e^{2 x}$ and

$0$ .

$f^{- 1} (3 e^{2 x} + 1) = \ln ({(e^{2 x})}^{\frac{1}{2}})$

Step 5.2.6.2

Multiply the exponents in

${(e^{2 x})}^{\frac{1}{2}}$ .

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Step 5.2.6.2.1

Apply the power rule and multiply exponents,

${(a^{m})}^{n} = a^{m n}$ .

$f^{- 1} (3 e^{2 x} + 1) = \ln (e^{2 x (\frac{1}{2})})$

Step 5.2.6.2.2

Cancel the common factor of

$2$ .

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Step 5.2.6.2.2.1

Factor

$2$ out of

$2 x$ .

$f^{- 1} (3 e^{2 x} + 1) = \ln (e^{2 (x) (\frac{1}{2})})$

Step 5.2.6.2.2.2

Cancel the common factor.

$f^{- 1} (3 e^{2 x} + 1) = \ln (e^{2 x (\frac{1}{2})})$

Step 5.2.6.2.2.3

Rewrite the expression.

$f^{- 1} (3 e^{2 x} + 1) = \ln (e^{x})$

Step 5.2.7

Use logarithm rules to move

$x$ out of the exponent.

$f^{- 1} (3 e^{2 x} + 1) = x \ln (e)$

Step 5.2.8

The natural logarithm of

$e$ is

$1$ .

$f^{- 1} (3 e^{2 x} + 1) = x \cdot 1$

Step 5.2.9

Multiply

$x$ by

$1$ .

$f^{- 1} (3 e^{2 x} + 1) = x$

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

$f (\frac{\ln (\frac{x}{3} - \frac{1}{3})}{2})$ by substituting in the value of

$f^{- 1}$ into

$f$ .

$f (\frac{\ln (\frac{x}{3} - \frac{1}{3})}{2}) = 3 e^{2 (\frac{\ln (\frac{x}{3} - \frac{1}{3})}{2})} + 1$

Step 5.3.3

Simplify each term.

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Step 5.3.3.1

Cancel the common factor of

$2$ .

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Step 5.3.3.1.1

Combine the numerators over the common denominator.

$f (\frac{\ln (\frac{x}{3} - \frac{1}{3})}{2}) = 3 e^{2 (\frac{\ln (\frac{x - 1}{3})}{2})} + 1$

Step 5.3.3.1.2

Cancel the common factor.

$f (\frac{\ln (\frac{x}{3} - \frac{1}{3})}{2}) = 3 e^{2 (\frac{\ln (\frac{x - 1}{3})}{2})} + 1$

Step 5.3.3.1.3

Rewrite the expression.

$f (\frac{\ln (\frac{x}{3} - \frac{1}{3})}{2}) = 3 e^{\ln (\frac{x - 1}{3})} + 1$

Step 5.3.3.2

Exponentiation and log are inverse functions.

$f (\frac{\ln (\frac{x}{3} - \frac{1}{3})}{2}) = 3 (\frac{x - 1}{3}) + 1$

Step 5.3.3.3

Cancel the common factor of

$3$ .

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Step 5.3.3.3.1

Cancel the common factor.

$f (\frac{\ln (\frac{x}{3} - \frac{1}{3})}{2}) = 3 (\frac{x - 1}{3}) + 1$

Step 5.3.3.3.2

Rewrite the expression.

$f (\frac{\ln (\frac{x}{3} - \frac{1}{3})}{2}) = x - 1 + 1$

Step 5.3.4

Combine the opposite terms in

$x - 1 + 1$ .

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Step 5.3.4.1

Add

$- 1$ and

$1$ .

$f (\frac{\ln (\frac{x}{3} - \frac{1}{3})}{2}) = x + 0$

Step 5.3.4.2

Add

$x$ and

$0$ .

$f (\frac{\ln (\frac{x}{3} - \frac{1}{3})}{2}) = x$

Step 5.4

Since

$f^{- 1} (f (x)) = x$ and

$f (f^{- 1} (x)) = x$ , then

$f^{- 1} (x) = \frac{\ln (\frac{x}{3} - \frac{1}{3})}{2}$ is the inverse of

$f (x) = 3 e^{2 x} + 1$ .

$f^{- 1} (x) = \frac{\ln (\frac{x}{3} - \frac{1}{3})}{2}$