Algebra Examples

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

Step 1

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

$y = - 2 x + 1$

Step 2

Interchange the variables.

$x = - 2 y + 1$

Step 3

Solve for $y$ .

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Step 3.1

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

$- 2 y + 1 = x$

Step 3.2

Subtract $1$ from both sides of the equation.

$- 2 y = x - 1$

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $- 2$ .

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Step 3.3.2.1.1

Cancel the common factor.

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

Step 3.3.2.1.2

Divide $y$ by $1$ .

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

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Simplify each term.

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Step 3.3.3.1.1

Move the negative in front of the fraction.

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

Step 3.3.3.1.2

Dividing two negative values results in a positive value.

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

Step 4

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

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

Step 5

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

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Step 5.1

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

Step 5.2

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

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Step 5.2.1

Set up the composite result function.

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

Step 5.2.2

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

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

Step 5.2.3

Combine the numerators over the common denominator.

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

Step 5.2.4

Simplify each term.

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Step 5.2.4.1

Apply the distributive property.

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

Step 5.2.4.2

Multiply $- 2$ by $- 1$ .

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

Step 5.2.4.3

Multiply $- 1$ by $1$ .

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

Step 5.2.5

Simplify terms.

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Step 5.2.5.1

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

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Step 5.2.5.1.1

Add $- 1$ and $1$ .

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

Step 5.2.5.1.2

Add $2 x$ and $0$ .

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

Step 5.2.5.2

Cancel the common factor of $2$ .

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Step 5.2.5.2.1

Cancel the common factor.

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

Step 5.2.5.2.2

Divide $x$ by $1$ .

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

Step 5.3

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

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Step 5.3.1

Set up the composite result function.

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

Step 5.3.2

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

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

Step 5.3.3

Simplify each term.

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Step 5.3.3.1

Apply the distributive property.

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

Step 5.3.3.2

Cancel the common factor of $2$ .

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Step 5.3.3.2.1

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

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

Step 5.3.3.2.2

Factor $2$ out of $- 2$ .

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

Step 5.3.3.2.3

Cancel the common factor.

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

Step 5.3.3.2.4

Rewrite the expression.

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

Step 5.3.3.3

Multiply $- 1$ by $- 1$ .

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

Step 5.3.3.4

Multiply $x$ by $1$ .

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

Step 5.3.3.5

Cancel the common factor of $2$ .

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Step 5.3.3.5.1

Factor $2$ out of $- 2$ .

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

Step 5.3.3.5.2

Cancel the common factor.

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

Step 5.3.3.5.3

Rewrite the expression.

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

Step 5.3.4

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

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Step 5.3.4.1

Add $- 1$ and $1$ .

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

Step 5.3.4.2

Add $x$ and $0$ .

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

Step 5.4

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

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