Find the Inverse 9/(x+3)

$\frac{9}{x + 3}$

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

Rewrite the equation as $\frac{9}{y + 3} = x$ .

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

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$y + 3, 1$

Remove parentheses.

$y + 3, 1$

The LCM of one and any expression is the expression.

$y + 3$

Multiply each term in $\frac{9}{y + 3} = x$ by $y + 3$ to eliminate the fractions.

Multiply each term in $\frac{9}{y + 3} = x$ by $y + 3$ .

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

Simplify the left side.

Cancel the common factor of $y + 3$ .

Cancel the common factor.

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

Rewrite the expression.

$9 = x (y + 3)$

Simplify the right side.

Apply the distributive property.

$9 = x y + x \cdot 3$

Move $3$ to the left of $x$ .

$9 = x y + 3 x$

Solve the equation.

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

$x y + 3 x = 9$

Subtract $3 x$ from both sides of the equation.

$x y = 9 - 3 x$

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

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

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

Simplify the left side.

Cancel the common factor of $x$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Cancel the common factor of $x$ .

Cancel the common factor.

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

Divide $- 3$ by $1$ .

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

Step 3

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

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

Step 4

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

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

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

Simplify each term.

Multiply the numerator by the reciprocal of the denominator.

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

Cancel the common factor of $9$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

Subtract $3$ from $3$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

Cancel the common factor of $9$ and $(\frac{9}{x} - 3) + 3$ .

Factor $3$ out of $9$ .

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

Cancel the common factors.

Factor $3$ out of $\frac{9}{x}$ .

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

Factor $3$ out of $- 3$ .

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

Factor $3$ out of $3 \frac{3}{x} + 3 \cdot - 1$ .

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

Factor $3$ out of $3$ .

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

Factor $3$ out of $3 (\frac{3}{x} - 1) + 3 (1)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the denominator.

Add $- 1$ and $1$ .

$f (\frac{9}{x} - 3) = \frac{3}{\frac{3}{x} + 0}$

Add $\frac{3}{x}$ and $0$ .

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

Multiply the numerator by the reciprocal of the denominator.

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Find the Inverse 9/(x+3)

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