Find the Inverse (x-3)/7

$\frac{x - 3}{7}$

Step 1

Interchange the variables.

$x = \frac{y - 3}{7}$

Step 2

Solve for $y$ .

Rewrite the equation as $\frac{y - 3}{7} = x$ .

$\frac{y - 3}{7} = x$

Multiply both sides of the equation by $7$ .

$7 \frac{y - 3}{7} = 7 x$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$7 \frac{y - 3}{7} = 7 x$

Rewrite the expression.

$y - 3 = 7 x$

Add $3$ to both sides of the equation.

$y = 7 x + 3$

Step 3

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

$f^{- 1} (x) = 7 x + 3$

Step 4

Verify if $f^{- 1} (x) = 7 x + 3$ is the inverse of $f (x) = \frac{x - 3}{7}$ .

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 $f^{- 1} (\frac{x - 3}{7})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{x - 3}{7}) = 7 (\frac{x - 3}{7}) + 3$

Cancel the common factor of $7$ .

Cancel the common factor.

$f^{- 1} (\frac{x - 3}{7}) = 7 (\frac{x - 3}{7}) + 3$

Rewrite the expression.

$f^{- 1} (\frac{x - 3}{7}) = x - 3 + 3$

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

Add $- 3$ and $3$ .

$f^{- 1} (\frac{x - 3}{7}) = x + 0$

Add $x$ and $0$ .

$f^{- 1} (\frac{x - 3}{7}) = x$

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

Set up the composite result function.

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

Evaluate $f (7 x + 3)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (7 x + 3) = \frac{(7 x + 3) - 3}{7}$

Simplify the numerator.

Subtract $3$ from $3$ .

$f (7 x + 3) = \frac{7 x + 0}{7}$

Add $7 x$ and $0$ .

$f (7 x + 3) = \frac{7 x}{7}$

Cancel the common factor of $7$ .

Cancel the common factor.

$f (7 x + 3) = \frac{7 x}{7}$

Divide $x$ by $1$ .

$f (7 x + 3) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = 7 x + 3$ is the inverse of $f (x) = \frac{x - 3}{7}$ .

$f^{- 1} (x) = 7 x + 3$

Find the Inverse (x-3)/7

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