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

$f (x) = - 2 x + 1$

Write $f (x) = - 2 x + 1$ as an equation.

$y = - 2 x + 1$

Interchange the variables.

$x = - 2 y + 1$

Solve for $y$ .

Rewrite the equation as $- 2 y + 1 = x$ .

$- 2 y + 1 = x$

Subtract $1$ from both sides of the equation.

$- 2 y = x - 1$

Divide each term in $- 2 y = x - 1$ by $- 2$ and simplify.

Divide each term in $- 2 y = x - 1$ by $- 2$ .

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

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

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

Dividing two negative values results in a positive value.

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

Simplify each term.

Apply the distributive property.

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

Multiply $- 2$ by $- 1$ .

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

Multiply $- 1$ by $1$ .

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

Simplify terms.

Combine the opposite terms in $2 x - 1 + 1$ .

Add $- 1$ and $1$ .

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

Add $2 x$ and $0$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

Simplify each term.

Apply the distributive property.

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

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{x}{2}$ into the numerator.

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

Factor $2$ out of $- 2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 1$ by $- 1$ .

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

Multiply $x$ by $1$ .

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

Cancel the common factor of $2$ .

Factor $2$ out of $- 2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Add $- 1$ and $1$ .

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

Add $x$ and $0$ .

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

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

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

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

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