Find the Inverse f(x)=1/9x+2

$f (x) = \frac{1}{9} x + 2$

Step 1

Write $f (x) = \frac{1}{9} x + 2$ as an equation.

$y = \frac{1}{9} x + 2$

Step 2

Interchange the variables.

$x = \frac{1}{9} y + 2$

Step 3

Solve for $y$ .

Rewrite the equation as $\frac{1}{9} y + 2 = x$ .

$\frac{1}{9} y + 2 = x$

Combine $\frac{1}{9}$ and $y$ .

$\frac{y}{9} + 2 = x$

Subtract $2$ from both sides of the equation.

$\frac{y}{9} = x - 2$

Multiply both sides of the equation by $9$ .

$9 \frac{y}{9} = 9 (x - 2)$

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $9$ .

Cancel the common factor.

$9 \frac{y}{9} = 9 (x - 2)$

Rewrite the expression.

$y = 9 (x - 2)$

Simplify the right side.

Simplify $9 (x - 2)$ .

Apply the distributive property.

$y = 9 x + 9 \cdot - 2$

Multiply $9$ by $- 2$ .

$y = 9 x - 18$

Step 4

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

$f^{- 1} (x) = 9 x - 18$

Step 5

Verify if $f^{- 1} (x) = 9 x - 18$ is the inverse of $f (x) = \frac{1}{9} x + 2$ .

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 $f^{- 1} (\frac{1}{9} x + 2)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{1}{9} x + 2) = 9 (\frac{1}{9} x + 2) - 18$

Simplify each term.

Combine $\frac{1}{9}$ and $x$ .

$f^{- 1} (\frac{1}{9} x + 2) = 9 (\frac{x}{9} + 2) - 18$

Apply the distributive property.

$f^{- 1} (\frac{1}{9} x + 2) = 9 (\frac{x}{9}) + 9 \cdot 2 - 18$

Cancel the common factor of $9$ .

Cancel the common factor.

$f^{- 1} (\frac{1}{9} x + 2) = 9 (\frac{x}{9}) + 9 \cdot 2 - 18$

Rewrite the expression.

$f^{- 1} (\frac{1}{9} x + 2) = x + 9 \cdot 2 - 18$

Multiply $9$ by $2$ .

$f^{- 1} (\frac{1}{9} x + 2) = x + 18 - 18$

Combine the opposite terms in $x + 18 - 18$ .

Subtract $18$ from $18$ .

$f^{- 1} (\frac{1}{9} x + 2) = x + 0$

Add $x$ and $0$ .

$f^{- 1} (\frac{1}{9} x + 2) = x$

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

Set up the composite result function.

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

Evaluate $f (9 x - 18)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (9 x - 18) = \frac{1}{9} \cdot (9 x - 18) + 2$

Simplify each term.

Apply the distributive property.

$f (9 x - 18) = \frac{1}{9} \cdot (9 x) + \frac{1}{9} \cdot - 18 + 2$

Cancel the common factor of $9$ .

Factor $9$ out of $9 x$ .

$f (9 x - 18) = \frac{1}{9} \cdot (9 (x)) + \frac{1}{9} \cdot - 18 + 2$

Cancel the common factor.

$f (9 x - 18) = \frac{1}{9} \cdot (9 x) + \frac{1}{9} \cdot - 18 + 2$

Rewrite the expression.

$f (9 x - 18) = x + \frac{1}{9} \cdot - 18 + 2$

Cancel the common factor of $9$ .

Factor $9$ out of $- 18$ .

$f (9 x - 18) = x + \frac{1}{9} \cdot (9 (- 2)) + 2$

Cancel the common factor.

$f (9 x - 18) = x + \frac{1}{9} \cdot (9 \cdot - 2) + 2$

Rewrite the expression.

$f (9 x - 18) = x - 2 + 2$

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

Add $- 2$ and $2$ .

$f (9 x - 18) = x + 0$

Add $x$ and $0$ .

$f (9 x - 18) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = 9 x - 18$ is the inverse of $f (x) = \frac{1}{9} x + 2$ .

$f^{- 1} (x) = 9 x - 18$

Find the Inverse f(x)=1/9x+2

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