Find the Inverse f(x)=(x-1)/2

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

Write $f (x) = \frac{x - 1}{2}$ as an equation.

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

Interchange the variables.

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

Solve for $y$ .

Rewrite the equation as $\frac{y - 1}{2} = x$ .

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

Multiply both sides of the equation by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

$y - 1 = 2 x$

Add $1$ to both sides of the equation.

$y = 2 x + 1$

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

Add $- 1$ and $1$ .

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

Add $x$ and $0$ .

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

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 $f$ .

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

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

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

Find the Inverse f(x)=(x-1)/2

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