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

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

Subtract $8$ from both sides of the equation.

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

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.

$y, 1, 1$

The LCM of one and any expression is the expression.

$y$

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

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

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

Simplify the left side.

Cancel the common factor of $y$ .

Cancel the common factor.

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

Rewrite the expression.

$1 = x y - 8 y$

Solve the equation.

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

$x y - 8 y = 1$

Factor $y$ out of $x y - 8 y$ .

Factor $y$ out of $x y$ .

$y x - 8 y = 1$

Factor $y$ out of $- 8 y$ .

$y x + y \cdot - 8 = 1$

Factor $y$ out of $y x + y \cdot - 8$ .

$y (x - 8) = 1$

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

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

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

Simplify the left side.

Cancel the common factor of $x - 8$ .

Cancel the common factor.

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

Divide $y$ 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 the denominator.

Subtract $8$ from $8$ .

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

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

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

Multiply the numerator by the reciprocal of the denominator.

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

Multiply $x$ by $1$ .

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

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 each term.

Multiply the numerator by the reciprocal of the denominator.

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

Multiply $x - 8$ by $1$ .

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

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

Add $- 8$ and $8$ .

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

Add $x$ and $0$ .

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

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

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