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 $y$ .

Tap for more steps...

Step 3.1

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

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

Step 3.2

Subtract $1$ from both sides of the equation.

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

Step 3.3

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

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

Step 3.4

Expand the left side.

Tap for more steps...

Step 3.4.1

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

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

Step 3.4.2

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

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

Step 3.4.3

Multiply $2$ by $1$ .

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

Step 3.5

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

Tap for more steps...

Step 3.5.1

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

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

Step 3.5.2

Simplify the left side.

Tap for more steps...

Step 3.5.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 3.5.2.1.1

Cancel the common factor.

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

Step 3.5.2.1.2

Divide $y$ by $1$ .

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

Step 4

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

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

Step 5

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

Tap for more steps...

Step 5.1

To verify the inverse, check if $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Step 5.2

Evaluate $f^{- 1} (f (x))$ .

Tap for more steps...

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 $f^{- 1}$ .

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

Step 5.2.3

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

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

Step 5.2.4

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

Tap for more steps...

Step 5.2.4.1

Subtract $1$ from $1$ .

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

Step 5.2.4.2

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

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

Step 5.2.5

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

Tap for more steps...

Step 5.2.5.1

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

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

Step 5.2.5.2

Cancel the common factor of $2$ .

Tap for more steps...

Step 5.2.5.2.1

Factor $2$ out of $2 x$ .

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

Step 5.2.5.2.2

Cancel the common factor.

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

Step 5.2.5.2.3

Rewrite the expression.

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

Step 5.2.6

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

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

Step 5.2.7

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

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

Step 5.2.8

Multiply $x$ by $1$ .

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

Step 5.3

Evaluate $f (f^{- 1} (x))$ .

Tap for more steps...

Step 5.3.1

Set up the composite result function.

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

Step 5.3.2

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

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

Step 5.3.3

Simplify each term.

Tap for more steps...

Step 5.3.3.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 5.3.3.1.1

Cancel the common factor.

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

Step 5.3.3.1.2

Rewrite the expression.

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

Step 5.3.3.2

Exponentiation and log are inverse functions.

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

Step 5.3.4

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

Tap for more steps...

Step 5.3.4.1

Add $- 1$ and $1$ .

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

Step 5.3.4.2

Add $x$ and $0$ .

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

Step 5.4

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

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