Find the Inverse 8/(3+x)

$\frac{8}{3 + x}$

Step 1

Interchange the variables.

$x = \frac{8}{3 + y}$

Step 2

Solve for $y$ .

Rewrite the equation as $\frac{8}{3 + y} = x$ .

$\frac{8}{3 + y} = x$

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.

$3 + y, 1$

Remove parentheses.

$3 + y, 1$

The LCM of one and any expression is the expression.

$3 + y$

Multiply each term in $\frac{8}{3 + y} = x$ by $3 + y$ to eliminate the fractions.

Multiply each term in $\frac{8}{3 + y} = x$ by $3 + y$ .

$\frac{8}{3 + y} (3 + y) = x (3 + y)$

Simplify the left side.

Cancel the common factor of $3 + y$ .

Cancel the common factor.

$\frac{8}{3 + y} (3 + y) = x (3 + y)$

Rewrite the expression.

$8 = x (3 + y)$

Simplify the right side.

Apply the distributive property.

$8 = x \cdot 3 + x y$

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

$8 = 3 x + x y$

Solve the equation.

Rewrite the equation as $3 x + x y = 8$ .

$3 x + x y = 8$

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

$x y = 8 - 3 x$

Divide each term in $x y = 8 - 3 x$ by $x$ and simplify.

Divide each term in $x y = 8 - 3 x$ by $x$ .

$\frac{x y}{x} = \frac{8}{x} + \frac{- 3 x}{x}$

Simplify the left side.

Cancel the common factor of $x$ .

Cancel the common factor.

$\frac{x y}{x} = \frac{8}{x} + \frac{- 3 x}{x}$

Divide $y$ by $1$ .

$y = \frac{8}{x} + \frac{- 3 x}{x}$

Simplify the right side.

Cancel the common factor of $x$ .

Cancel the common factor.

$y = \frac{8}{x} + \frac{- 3 x}{x}$

Divide $- 3$ by $1$ .

$y = \frac{8}{x} - 3$

Step 3

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

$f^{- 1} (x) = \frac{8}{x} - 3$

Step 4

Verify if $f^{- 1} (x) = \frac{8}{x} - 3$ is the inverse of $f (x) = \frac{8}{3 + x}$ .

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 $f^{- 1} (\frac{8}{3 + x})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{8}{3 + x}) = \frac{8}{\frac{8}{3 + x}} - 3$

Simplify each term.

Multiply the numerator by the reciprocal of the denominator.

$f^{- 1} (\frac{8}{3 + x}) = 8 (\frac{3 + x}{8}) - 3$

Cancel the common factor of $8$ .

Cancel the common factor.

$f^{- 1} (\frac{8}{3 + x}) = 8 (\frac{3 + x}{8}) - 3$

Rewrite the expression.

$f^{- 1} (\frac{8}{3 + x}) = 3 + x - 3$

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

Subtract $3$ from $3$ .

$f^{- 1} (\frac{8}{3 + x}) = x + 0$

Add $x$ and $0$ .

$f^{- 1} (\frac{8}{3 + x}) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{8}{x} - 3)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{8}{x} - 3) = \frac{8}{3 + \frac{8}{x} - 3}$

Remove parentheses.

$f (\frac{8}{x} - 3) = \frac{8}{3 + \frac{8}{x} - 3}$

Simplify the denominator.

Subtract $3$ from $3$ .

$f (\frac{8}{x} - 3) = \frac{8}{\frac{8}{x} + 0}$

Add $\frac{8}{x}$ and $0$ .

$f (\frac{8}{x} - 3) = \frac{8}{\frac{8}{x}}$

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{8}{x} - 3) = 8 (\frac{x}{8})$

Cancel the common factor of $8$ .

Cancel the common factor.

$f (\frac{8}{x} - 3) = 8 (\frac{x}{8})$

Rewrite the expression.

$f (\frac{8}{x} - 3) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{8}{x} - 3$ is the inverse of $f (x) = \frac{8}{3 + x}$ .

$f^{- 1} (x) = \frac{8}{x} - 3$

Find the Inverse 8/(3+x)

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