Trigonometry Examples

$f (x) = 2 e^{x - 2} + 4$

Step 1

Write $f (x) = 2 e^{x - 2} + 4$ as an equation.

$y = 2 e^{x - 2} + 4$

Step 2

Interchange the variables.

$x = 2 e^{y - 2} + 4$

Step 3

Solve for $y$ .

Tap for more steps...

Step 3.1

Rewrite the equation as $2 e^{y - 2} + 4 = x$ .

$2 e^{y - 2} + 4 = x$

Step 3.2

Subtract $4$ from both sides of the equation.

$2 e^{y - 2} = x - 4$

Step 3.3

Divide each term in $2 e^{y - 2} = x - 4$ by $2$ and simplify.

Tap for more steps...

Step 3.3.1

Divide each term in $2 e^{y - 2} = x - 4$ by $2$ .

$\frac{2 e^{y - 2}}{2} = \frac{x}{2} + \frac{- 4}{2}$

Step 3.3.2

Simplify the left side.

Tap for more steps...

Step 3.3.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 3.3.2.1.1

Cancel the common factor.

$\frac{2 e^{y - 2}}{2} = \frac{x}{2} + \frac{- 4}{2}$

Step 3.3.2.1.2

Divide $e^{y - 2}$ by $1$ .

$e^{y - 2} = \frac{x}{2} + \frac{- 4}{2}$

Step 3.3.3

Simplify the right side.

Tap for more steps...

Step 3.3.3.1

Divide $- 4$ by $2$ .

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

Step 3.4

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

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

Step 3.5

Expand the left side.

Tap for more steps...

Step 3.5.1

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

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

Step 3.5.2

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

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

Step 3.5.3

Multiply $y - 2$ by $1$ .

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

Step 3.6

Add $2$ to both sides of the equation.

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

Step 4

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

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

Step 5

Verify if $f^{- 1} (x) = \ln (\frac{x}{2} - 2) + 2$ is the inverse of $f (x) = 2 e^{x - 2} + 4$ .

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 $f^{- 1} (2 e^{x - 2} + 4)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (2 e^{x - 2} + 4) = \ln (\frac{2 e^{x - 2} + 4}{2} - 2) + 2$

Step 5.2.3

Simplify each term.

Tap for more steps...

Step 5.2.3.1

Cancel the common factor of $2 e^{x - 2} + 4$ and $2$ .

Tap for more steps...

Step 5.2.3.1.1

Factor $2$ out of $2 e^{x - 2}$ .

$f^{- 1} (2 e^{x - 2} + 4) = \ln (\frac{2 (e^{x - 2}) + 4}{2} - 2) + 2$

Step 5.2.3.1.2

Factor $2$ out of $4$ .

$f^{- 1} (2 e^{x - 2} + 4) = \ln (\frac{2 (e^{x - 2}) + 2 \cdot 2}{2} - 2) + 2$

Step 5.2.3.1.3

Factor $2$ out of $2 (e^{x - 2}) + 2 (2)$ .

$f^{- 1} (2 e^{x - 2} + 4) = \ln (\frac{2 (e^{x - 2} + 2)}{2} - 2) + 2$

Step 5.2.3.1.4

Cancel the common factors.

Tap for more steps...

Step 5.2.3.1.4.1

Factor $2$ out of $2$ .

$f^{- 1} (2 e^{x - 2} + 4) = \ln (\frac{2 (e^{x - 2} + 2)}{2 (1)} - 2) + 2$

Step 5.2.3.1.4.2

Cancel the common factor.

$f^{- 1} (2 e^{x - 2} + 4) = \ln (\frac{2 (e^{x - 2} + 2)}{2 \cdot 1} - 2) + 2$

Step 5.2.3.1.4.3

Rewrite the expression.

$f^{- 1} (2 e^{x - 2} + 4) = \ln (\frac{e^{x - 2} + 2}{1} - 2) + 2$

Step 5.2.3.1.4.4

Divide $e^{x - 2} + 2$ by $1$ .

$f^{- 1} (2 e^{x - 2} + 4) = \ln (e^{x - 2} + 2 - 2) + 2$

Step 5.2.3.2

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

Tap for more steps...

Step 5.2.3.2.1

Subtract $2$ from $2$ .

$f^{- 1} (2 e^{x - 2} + 4) = \ln (e^{x - 2} + 0) + 2$

Step 5.2.3.2.2

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

$f^{- 1} (2 e^{x - 2} + 4) = \ln (e^{x - 2}) + 2$

Step 5.2.3.3

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

$f^{- 1} (2 e^{x - 2} + 4) = (x - 2) \ln (e) + 2$

Step 5.2.3.4

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

$f^{- 1} (2 e^{x - 2} + 4) = (x - 2) \cdot 1 + 2$

Step 5.2.3.5

Multiply $x - 2$ by $1$ .

$f^{- 1} (2 e^{x - 2} + 4) = x - 2 + 2$

Step 5.2.4

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

Tap for more steps...

Step 5.2.4.1

Add $- 2$ and $2$ .

$f^{- 1} (2 e^{x - 2} + 4) = x + 0$

Step 5.2.4.2

Add $x$ and $0$ .

$f^{- 1} (2 e^{x - 2} + 4) = x$

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

$f (\ln (\frac{x}{2} - 2) + 2) = 2 e^{(\ln (\frac{x}{2} - 2) + 2) - 2} + 4$

Step 5.3.3

Combine the opposite terms in $2 e^{(\ln (\frac{x}{2} - 2) + 2) - 2} + 4$ .

Tap for more steps...

Step 5.3.3.1

Subtract $2$ from $2$ .

$f (\ln (\frac{x}{2} - 2) + 2) = 2 e^{\ln (\frac{x}{2} - 2) + 0} + 4$

Step 5.3.3.2

Add $\ln (\frac{x}{2} - 2)$ and $0$ .

$f (\ln (\frac{x}{2} - 2) + 2) = 2 e^{\ln (\frac{x}{2} - 2)} + 4$

Step 5.3.4

Simplify each term.

Tap for more steps...

Step 5.3.4.1

Exponentiation and log are inverse functions.

$f (\ln (\frac{x}{2} - 2) + 2) = 2 (\frac{x}{2} - 2) + 4$

Step 5.3.4.2

Apply the distributive property.

$f (\ln (\frac{x}{2} - 2) + 2) = 2 (\frac{x}{2}) + 2 \cdot - 2 + 4$

Step 5.3.4.3

Cancel the common factor of $2$ .

Tap for more steps...

Step 5.3.4.3.1

Cancel the common factor.

$f (\ln (\frac{x}{2} - 2) + 2) = 2 (\frac{x}{2}) + 2 \cdot - 2 + 4$

Step 5.3.4.3.2

Rewrite the expression.

$f (\ln (\frac{x}{2} - 2) + 2) = x + 2 \cdot - 2 + 4$

Step 5.3.4.4

Multiply $2$ by $- 2$ .

$f (\ln (\frac{x}{2} - 2) + 2) = x - 4 + 4$

Step 5.3.5

Combine the opposite terms in $x - 4 + 4$ .

Tap for more steps...

Step 5.3.5.1

Add $- 4$ and $4$ .

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

Step 5.3.5.2

Add $x$ and $0$ .

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

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \ln (\frac{x}{2} - 2) + 2$ is the inverse of $f (x) = 2 e^{x - 2} + 4$ .

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