Find the Inverse f(x)=(2x-4)

$f (x) = (2 x - 4)$

Step 1

Write $f (x) = (2 x - 4)$ as an equation.

$y = (2 x - 4)$

Step 2

Interchange the variables.

$x = 2 y - 4$

Step 3

Solve for $y$ .

Rewrite the equation as $2 y - 4 = x$ .

$2 y - 4 = x$

Add $4$ to both sides of the equation.

$2 y = x + 4$

Divide each term in $2 y = x + 4$ by $2$ and simplify.

Divide each term in $2 y = x + 4$ by $2$ .

$\frac{2 y}{2} = \frac{x}{2} + \frac{4}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y}{2} = \frac{x}{2} + \frac{4}{2}$

Divide $y$ by $1$ .

$y = \frac{x}{2} + \frac{4}{2}$

Simplify the right side.

Divide $4$ by $2$ .

$y = \frac{x}{2} + 2$

Step 4

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

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

Step 5

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

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

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

Set up the composite result function.

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

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

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

Cancel the common factor of $2 x - 4$ and $2$ .

Factor $2$ out of $2 x$ .

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

Factor $2$ out of $- 4$ .

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

Factor $2$ out of $2 (x) + 2 (- 2)$ .

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

Cancel the common factors.

Factor $2$ out of $2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $x - 2$ by $1$ .

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

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

Add $- 2$ and $2$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

Remove parentheses.

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

Simplify each term.

Apply the distributive property.

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $2$ by $2$ .

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

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

Subtract $4$ from $4$ .

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

Add $x$ and $0$ .

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

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

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

Find the Inverse f(x)=(2x-4)

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