Find the Inverse 1/7x-9

$\frac{1}{7} x - 9$

Step 1

Interchange the variables.

$x = \frac{1}{7} y - 9$

Step 2

Solve for $y$ .

Rewrite the equation as $\frac{1}{7} y - 9 = x$ .

$\frac{1}{7} y - 9 = x$

Combine $\frac{1}{7}$ and $y$ .

$\frac{y}{7} - 9 = x$

Add $9$ to both sides of the equation.

$\frac{y}{7} = x + 9$

Multiply both sides of the equation by $7$ .

$7 \frac{y}{7} = 7 (x + 9)$

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$7 \frac{y}{7} = 7 (x + 9)$

Rewrite the expression.

$y = 7 (x + 9)$

Simplify the right side.

Simplify $7 (x + 9)$ .

Apply the distributive property.

$y = 7 x + 7 \cdot 9$

Multiply $7$ by $9$ .

$y = 7 x + 63$

Step 3

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

$f^{- 1} (x) = 7 x + 63$

Step 4

Verify if $f^{- 1} (x) = 7 x + 63$ is the inverse of $f (x) = \frac{1}{7} x - 9$ .

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 $f^{- 1} (\frac{1}{7} x - 9)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{1}{7} x - 9) = 7 (\frac{1}{7} x - 9) + 63$

Simplify each term.

Combine $\frac{1}{7}$ and $x$ .

$f^{- 1} (\frac{1}{7} x - 9) = 7 (\frac{x}{7} - 9) + 63$

Apply the distributive property.

$f^{- 1} (\frac{1}{7} x - 9) = 7 (\frac{x}{7}) + 7 \cdot - 9 + 63$

Cancel the common factor of $7$ .

Cancel the common factor.

$f^{- 1} (\frac{1}{7} x - 9) = 7 (\frac{x}{7}) + 7 \cdot - 9 + 63$

Rewrite the expression.

$f^{- 1} (\frac{1}{7} x - 9) = x + 7 \cdot - 9 + 63$

Multiply $7$ by $- 9$ .

$f^{- 1} (\frac{1}{7} x - 9) = x - 63 + 63$

Combine the opposite terms in $x - 63 + 63$ .

Add $- 63$ and $63$ .

$f^{- 1} (\frac{1}{7} x - 9) = x + 0$

Add $x$ and $0$ .

$f^{- 1} (\frac{1}{7} x - 9) = x$

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

Set up the composite result function.

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

Evaluate $f (7 x + 63)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (7 x + 63) = \frac{1}{7} \cdot (7 x + 63) - 9$

Simplify each term.

Apply the distributive property.

$f (7 x + 63) = \frac{1}{7} \cdot (7 x) + \frac{1}{7} \cdot 63 - 9$

Cancel the common factor of $7$ .

Factor $7$ out of $7 x$ .

$f (7 x + 63) = \frac{1}{7} \cdot (7 (x)) + \frac{1}{7} \cdot 63 - 9$

Cancel the common factor.

$f (7 x + 63) = \frac{1}{7} \cdot (7 x) + \frac{1}{7} \cdot 63 - 9$

Rewrite the expression.

$f (7 x + 63) = x + \frac{1}{7} \cdot 63 - 9$

Cancel the common factor of $7$ .

Factor $7$ out of $63$ .

$f (7 x + 63) = x + \frac{1}{7} \cdot (7 (9)) - 9$

Cancel the common factor.

$f (7 x + 63) = x + \frac{1}{7} \cdot (7 \cdot 9) - 9$

Rewrite the expression.

$f (7 x + 63) = x + 9 - 9$

Combine the opposite terms in $x + 9 - 9$ .

Subtract $9$ from $9$ .

$f (7 x + 63) = x + 0$

Add $x$ and $0$ .

$f (7 x + 63) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = 7 x + 63$ is the inverse of $f (x) = \frac{1}{7} x - 9$ .

$f^{- 1} (x) = 7 x + 63$

Find the Inverse 1/7x-9

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