Find the Inverse (x-3)/4

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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

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

Multiply both sides of the equation by $4$ .

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

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

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

Rewrite the expression.

$y - 3 = 4 x$

Add $3$ to both sides of the equation.

$y = 4 x + 3$

Step 3

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

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

Step 4

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

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

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

Cancel the common factor of $4$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

Add $- 3$ and $3$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

Simplify the numerator.

Subtract $3$ from $3$ .

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

Add $4 x$ and $0$ .

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

Cancel the common factor of $4$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

$f (4 x + 3) = x$

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

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

Find the Inverse (x-3)/4

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