Find the Inverse (2x-2)/4

$\frac{2 x - 2}{4}$

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

Rewrite the equation as $\frac{2 y - 2}{4} = x$ .

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

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

Factor $2$ out of $2 y$ .

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

Factor $2$ out of $- 2$ .

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

Factor $2$ out of $2 (y) + 2 (- 1)$ .

$\frac{2 (y - 1)}{4} = x$

Cancel the common factors.

Factor $2$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

$\frac{y - 1}{2} = x$

Multiply both sides by $2$ .

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

Simplify.

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

$y - 1 = x \cdot 2$

Simplify the right side.

Move $2$ to the left of $x$ .

$y - 1 = 2 x$

Add $1$ to both sides of the equation.

$y = 2 x + 1$

Step 3

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

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

Step 4

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

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

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

Simplify each term.

Factor $2$ out of $2 x - 2$ .

Factor $2$ out of $2 x$ .

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

Factor $2$ out of $- 2$ .

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

Factor $2$ out of $2 (x) + 2 (- 1)$ .

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

Cancel the common factor of $2$ .

Factor $2$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x - 1$ by $1$ .

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

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

Add $- 1$ and $1$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

Cancel the common factor of $2 (2 x + 1) - 2$ and $4$ .

Factor $2$ out of $- 2$ .

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

Factor $2$ out of $2 (2 x + 1) + 2 (- 1)$ .

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

Cancel the common factors.

Factor $2$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the numerator.

Subtract $1$ from $1$ .

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

Add $2 x$ and $0$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

$f (2 x + 1) = x$

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

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

Find the Inverse (2x-2)/4

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