Find the Inverse f(x)=3x+9

$f (x) = 3 x + 9$

Write $f (x) = 3 x + 9$ as an equation.

$y = 3 x + 9$

Interchange the variables.

$x = 3 y + 9$

Solve for $y$ .

Rewrite the equation as $3 y + 9 = x$ .

$3 y + 9 = x$

Subtract $9$ from both sides of the equation.

$3 y = x - 9$

Divide each term in $3 y = x - 9$ by $3$ and simplify.

Divide each term in $3 y = x - 9$ by $3$ .

$\frac{3 y}{3} = \frac{x}{3} + \frac{- 9}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 y}{3} = \frac{x}{3} + \frac{- 9}{3}$

Divide $y$ by $1$ .

$y = \frac{x}{3} + \frac{- 9}{3}$

Simplify the right side.

Divide $- 9$ by $3$ .

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

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

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

Verify if $f^{- 1} (x) = \frac{x}{3} - 3$ is the inverse of $f (x) = 3 x + 9$ .

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 $f^{- 1} (3 x + 9)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (3 x + 9) = \frac{3 x + 9}{3} - 3$

Cancel the common factor of $3 x + 9$ and $3$ .

Factor $3$ out of $3 x$ .

$f^{- 1} (3 x + 9) = \frac{3 (x) + 9}{3} - 3$

Factor $3$ out of $9$ .

$f^{- 1} (3 x + 9) = \frac{3 (x) + 3 \cdot 3}{3} - 3$

Factor $3$ out of $3 (x) + 3 (3)$ .

$f^{- 1} (3 x + 9) = \frac{3 (x + 3)}{3} - 3$

Cancel the common factors.

Factor $3$ out of $3$ .

$f^{- 1} (3 x + 9) = \frac{3 (x + 3)}{3 (1)} - 3$

Cancel the common factor.

$f^{- 1} (3 x + 9) = \frac{3 (x + 3)}{3 \cdot 1} - 3$

Rewrite the expression.

$f^{- 1} (3 x + 9) = \frac{x + 3}{1} - 3$

Divide $x + 3$ by $1$ .

$f^{- 1} (3 x + 9) = x + 3 - 3$

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

Subtract $3$ from $3$ .

$f^{- 1} (3 x + 9) = x + 0$

Add $x$ and $0$ .

$f^{- 1} (3 x + 9) = x$

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

Set up the composite result function.

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

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

$f (\frac{x}{3} - 3) = 3 (\frac{x}{3} - 3) + 9$

Simplify each term.

Apply the distributive property.

$f (\frac{x}{3} - 3) = 3 (\frac{x}{3}) + 3 \cdot - 3 + 9$

Cancel the common factor of $3$ .

Cancel the common factor.

$f (\frac{x}{3} - 3) = 3 (\frac{x}{3}) + 3 \cdot - 3 + 9$

Rewrite the expression.

$f (\frac{x}{3} - 3) = x + 3 \cdot - 3 + 9$

Multiply $3$ by $- 3$ .

$f (\frac{x}{3} - 3) = x - 9 + 9$

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

Add $- 9$ and $9$ .

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

Add $x$ and $0$ .

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

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{x}{3} - 3$ is the inverse of $f (x) = 3 x + 9$ .

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

Find the Inverse f(x)=3x+9

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