Find the Inverse y=1/4x+1/2

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

Rewrite the equation as $\frac{1}{4} y + \frac{1}{2} = x$ .

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

Combine $\frac{1}{4}$ and $y$ .

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

Subtract $\frac{1}{2}$ from both sides of the equation.

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

Multiply both sides of the equation by $4$ .

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

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the right side.

Simplify $4 (x - \frac{1}{2})$ .

Apply the distributive property.

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

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{2}$ into the numerator.

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

Factor $2$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

$y = 4 x + 2 \cdot - 1$

Multiply $2$ by $- 1$ .

$y = 4 x - 2$

Step 3

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

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

Step 4

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

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

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

Simplify each term.

Combine $\frac{1}{4}$ and $x$ .

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

Apply the distributive property.

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

Cancel the common factor of $4$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $2$ .

Factor $2$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Subtract $2$ from $2$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

Simplify each term.

Apply the distributive property.

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

Cancel the common factor of $4$ .

Factor $4$ out of $4 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $2$ .

Factor $2$ out of $4$ .

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

Factor $2$ out of $- 2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Combine $\frac{1}{2}$ and $- 1$ .

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

Move the negative in front of the fraction.

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

Combine the opposite terms in $x - \frac{1}{2} + \frac{1}{2}$ .

Add $- \frac{1}{2}$ and $\frac{1}{2}$ .

$f (4 x - 2) = x + 0$

Add $x$ and $0$ .

$f (4 x - 2) = x$

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

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

Find the Inverse y=1/4x+1/2

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