Find the Inverse 1/4x-3

$\frac{1}{4} x - 3$

Step 1

Interchange the variables.

$x = \frac{1}{4} y - 3$

Step 2

Solve for $y$ .

Rewrite the equation as $\frac{1}{4} y - 3 = x$ .

$\frac{1}{4} y - 3 = x$

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

$\frac{y}{4} - 3 = x$

Add $3$ to both sides of the equation.

$\frac{y}{4} = x + 3$

Multiply both sides of the equation by $4$ .

$4 \frac{y}{4} = 4 (x + 3)$

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$4 \frac{y}{4} = 4 (x + 3)$

Rewrite the expression.

$y = 4 (x + 3)$

Simplify the right side.

Simplify $4 (x + 3)$ .

Apply the distributive property.

$y = 4 x + 4 \cdot 3$

Multiply $4$ by $3$ .

$y = 4 x + 12$

Step 3

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

$f^{- 1} (x) = 4 x + 12$

Step 4

Verify if $f^{- 1} (x) = 4 x + 12$ is the inverse of $f (x) = \frac{1}{4} x - 3$ .

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 $f^{- 1} (\frac{1}{4} x - 3)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{1}{4} x - 3) = 4 (\frac{1}{4} x - 3) + 12$

Simplify each term.

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

$f^{- 1} (\frac{1}{4} x - 3) = 4 (\frac{x}{4} - 3) + 12$

Apply the distributive property.

$f^{- 1} (\frac{1}{4} x - 3) = 4 (\frac{x}{4}) + 4 \cdot - 3 + 12$

Cancel the common factor of $4$ .

Cancel the common factor.

$f^{- 1} (\frac{1}{4} x - 3) = 4 (\frac{x}{4}) + 4 \cdot - 3 + 12$

Rewrite the expression.

$f^{- 1} (\frac{1}{4} x - 3) = x + 4 \cdot - 3 + 12$

Multiply $4$ by $- 3$ .

$f^{- 1} (\frac{1}{4} x - 3) = x - 12 + 12$

Combine the opposite terms in $x - 12 + 12$ .

Add $- 12$ and $12$ .

$f^{- 1} (\frac{1}{4} x - 3) = x + 0$

Add $x$ and $0$ .

$f^{- 1} (\frac{1}{4} x - 3) = x$

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

Set up the composite result function.

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

Evaluate $f (4 x + 12)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (4 x + 12) = \frac{1}{4} \cdot (4 x + 12) - 3$

Simplify each term.

Apply the distributive property.

$f (4 x + 12) = \frac{1}{4} \cdot (4 x) + \frac{1}{4} \cdot 12 - 3$

Cancel the common factor of $4$ .

Factor $4$ out of $4 x$ .

$f (4 x + 12) = \frac{1}{4} \cdot (4 (x)) + \frac{1}{4} \cdot 12 - 3$

Cancel the common factor.

$f (4 x + 12) = \frac{1}{4} \cdot (4 x) + \frac{1}{4} \cdot 12 - 3$

Rewrite the expression.

$f (4 x + 12) = x + \frac{1}{4} \cdot 12 - 3$

Cancel the common factor of $4$ .

Factor $4$ out of $12$ .

$f (4 x + 12) = x + \frac{1}{4} \cdot (4 (3)) - 3$

Cancel the common factor.

$f (4 x + 12) = x + \frac{1}{4} \cdot (4 \cdot 3) - 3$

Rewrite the expression.

$f (4 x + 12) = x + 3 - 3$

Combine the opposite terms in $x + 3 - 3$ .

Subtract $3$ from $3$ .

$f (4 x + 12) = x + 0$

Add $x$ and $0$ .

$f (4 x + 12) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = 4 x + 12$ is the inverse of $f (x) = \frac{1}{4} x - 3$ .

$f^{- 1} (x) = 4 x + 12$

Find the Inverse 1/4x-3

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