Algebra Examples

g (x) = \frac{1}{x - 1}

$g (x) = \frac{1}{x - 1}$

Step 1

Write

g (x) = \frac{1}{x - 1}

$g (x) = \frac{1}{x - 1}$ as an equation.

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

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

Step 2

Interchange the variables.

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

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

Step 3

Solve for

y

$y$ .

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Step 3.1

Rewrite the equation as

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

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

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

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

Step 3.2

Find the LCD of the terms in the equation.

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Step 3.2.1

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

y - 1, 1

$y - 1, 1$

Step 3.2.2

Remove parentheses.

y - 1, 1

$y - 1, 1$

Step 3.2.3

The LCM of one and any expression is the expression.

y - 1

$y - 1$

y - 1

$y - 1$

Step 3.3

Multiply each term in

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

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

y - 1

$y - 1$ to eliminate the fractions.

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Step 3.3.1

Multiply each term in

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

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

y - 1

$y - 1$ .

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

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

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of

y - 1

$y - 1$ .

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Step 3.3.2.1.1

Cancel the common factor.

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

Step 3.3.2.1.2

Rewrite the expression.

$1 = x (y - 1)$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Simplify by multiplying through.

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Step 3.3.3.1.1

Apply the distributive property.

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

Step 3.3.3.1.2

Move

$- 1$ to the left of

$x$ .

$1 = x y - 1 \cdot x$

Step 3.3.3.2

Rewrite

$- 1 x$ as

$- x$ .

$1 = x y - x$

Step 3.4

Solve the equation.

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Step 3.4.1

Rewrite the equation as

$x y - x = 1$ .

$x y - x = 1$

Step 3.4.2

Add

$x$ to both sides of the equation.

$x y = 1 + x$

Step 3.4.3

Divide each term in

$x y = 1 + x$ by

$x$ and simplify.

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Step 3.4.3.1

Divide each term in

$x y = 1 + x$ by

$x$ .

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

Step 3.4.3.2

Simplify the left side.

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Step 3.4.3.2.1

Cancel the common factor of

$x$ .

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Step 3.4.3.2.1.1

Cancel the common factor.

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

Step 3.4.3.2.1.2

Divide

$y$ by

$1$ .

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

Step 3.4.3.3

Simplify the right side.

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Step 3.4.3.3.1

Cancel the common factor of

$x$ .

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Step 3.4.3.3.1.1

Cancel the common factor.

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

Step 3.4.3.3.1.2

Rewrite the expression.

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

Step 4

Replace

$y$ with

$g^{- 1} (x)$ to show the final answer.

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

Step 5

Verify if

$g^{- 1} (x) = \frac{1}{x} + 1$ is the inverse of

$g (x) = \frac{1}{x - 1}$ .

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Step 5.1

To verify the inverse, check if

$g^{- 1} (g (x)) = x$ and

$g (g^{- 1} (x)) = x$ .

Step 5.2

Evaluate

$g^{- 1} (g (x))$ .

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Step 5.2.1

Set up the composite result function.

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

Step 5.2.2

Evaluate

$g^{- 1} (\frac{1}{x - 1})$ by substituting in the value of

$g$ into

$g^{- 1}$ .

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

Step 5.2.3

Simplify each term.

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Step 5.2.3.1

Multiply the numerator by the reciprocal of the denominator.

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

Step 5.2.3.2

Multiply

$x - 1$ by

$1$ .

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

Step 5.2.4

Combine the opposite terms in

$x - 1 + 1$ .

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Step 5.2.4.1

Add

$- 1$ and

$1$ .

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

Step 5.2.4.2

Add

$x$ and

$0$ .

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

Step 5.3

Evaluate

$g (g^{- 1} (x))$ .

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Step 5.3.1

Set up the composite result function.

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

Step 5.3.2

Evaluate

$g (\frac{1}{x} + 1)$ by substituting in the value of

$g^{- 1}$ into

$g$ .

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

Step 5.3.3

Simplify the denominator.

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Step 5.3.3.1

Subtract

$1$ from

$1$ .

$g (\frac{1}{x} + 1) = \frac{1}{\frac{1}{x} + 0}$

Step 5.3.3.2

Add

$\frac{1}{x}$ and

$0$ .

$g (\frac{1}{x} + 1) = \frac{1}{\frac{1}{x}}$

Step 5.3.4

Multiply the numerator by the reciprocal of the denominator.

$g (\frac{1}{x} + 1) = 1 x$

Step 5.3.5

Multiply

$x$ by

$1$ .

$g (\frac{1}{x} + 1) = x$

Step 5.4

Since

$g^{- 1} (g (x)) = x$ and

$g (g^{- 1} (x)) = x$ , then

$g^{- 1} (x) = \frac{1}{x} + 1$ is the inverse of

$g (x) = \frac{1}{x - 1}$ .

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