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

$f (x) = 3 x + 3$

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

$y = 3 x + 3$

Interchange the variables.

$x = 3 y + 3$

Solve for $y$ .

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

$3 y + 3 = x$

Subtract $3$ from both sides of the equation.

$3 y = x - 3$

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

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

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Divide $- 3$ by $3$ .

$y = \frac{x}{3} - 1$

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

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

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

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

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

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

Factor $3$ out of $3 x$ .

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

Factor $3$ out of $3$ .

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

Factor $3$ out of $3 (x) + 3 (1)$ .

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

Cancel the common factors.

Factor $3$ out of $3$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $x + 1$ by $1$ .

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

Combine the opposite terms in $x + 1 - 1$ .

Subtract $1$ from $1$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

Simplify each term.

Apply the distributive property.

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $3$ by $- 1$ .

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

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

Add $- 3$ and $3$ .

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

Add $x$ and $0$ .

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

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

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

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

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