Find the Inverse 1/(x+8)

$\frac{1}{x + 8}$

Step 1

Interchange the variables.

$x = \frac{1}{y + 8}$

Step 2

Solve for $y$ .

Rewrite the equation as $\frac{1}{y + 8} = x$ .

$\frac{1}{y + 8} = x$

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.

$y + 8, 1$

Remove parentheses.

$y + 8, 1$

The LCM of one and any expression is the expression.

$y + 8$

Multiply each term in $\frac{1}{y + 8} = x$ by $y + 8$ to eliminate the fractions.

Multiply each term in $\frac{1}{y + 8} = x$ by $y + 8$ .

$\frac{1}{y + 8} (y + 8) = x (y + 8)$

Simplify the left side.

Cancel the common factor of $y + 8$ .

Cancel the common factor.

$\frac{1}{y + 8} (y + 8) = x (y + 8)$

Rewrite the expression.

$1 = x (y + 8)$

Simplify the right side.

Apply the distributive property.

$1 = x y + x \cdot 8$

Move $8$ to the left of $x$ .

$1 = x y + 8 x$

Solve the equation.

Rewrite the equation as $x y + 8 x = 1$ .

$x y + 8 x = 1$

Subtract $8 x$ from both sides of the equation.

$x y = 1 - 8 x$

Divide each term in $x y = 1 - 8 x$ by $x$ and simplify.

Divide each term in $x y = 1 - 8 x$ by $x$ .

$\frac{x y}{x} = \frac{1}{x} + \frac{- 8 x}{x}$

Simplify the left side.

Cancel the common factor of $x$ .

Cancel the common factor.

$\frac{x y}{x} = \frac{1}{x} + \frac{- 8 x}{x}$

Divide $y$ by $1$ .

$y = \frac{1}{x} + \frac{- 8 x}{x}$

Simplify the right side.

Cancel the common factor of $x$ .

Cancel the common factor.

$y = \frac{1}{x} + \frac{- 8 x}{x}$

Divide $- 8$ by $1$ .

$y = \frac{1}{x} - 8$

Step 3

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

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

Step 4

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

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

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

Simplify each term.

Multiply the numerator by the reciprocal of the denominator.

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

Multiply $x + 8$ by $1$ .

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

Combine the opposite terms in $x + 8 - 8$ .

Subtract $8$ from $8$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

Simplify the denominator.

Add $- 8$ and $8$ .

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

Add $\frac{1}{x}$ and $0$ .

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

Multiply the numerator by the reciprocal of the denominator.

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

Multiply $x$ by $1$ .

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

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

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

Find the Inverse 1/(x+8)

Related Questions

Support LCMGCF.com by disabling your adblocker.