Algebra Examples

f (x) = 2 x + 1

$f (x) = 2 x + 1$

Step 1

Write

f (x) = 2 x + 1

$f (x) = 2 x + 1$ as an equation.

y = 2 x + 1

$y = 2 x + 1$

Step 2

Interchange the variables.

x = 2 y + 1

$x = 2 y + 1$

Step 3

Solve for

y

$y$ .

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Step 3.1

Rewrite the equation as

2 y + 1 = x

$2 y + 1 = x$ .

2 y + 1 = x

$2 y + 1 = x$

Step 3.2

Subtract

1

$1$ from both sides of the equation.

2 y = x - 1

$2 y = x - 1$

Step 3.3

Divide each term in

2 y = x - 1

$2 y = x - 1$ by

2

$2$ and simplify.

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Step 3.3.1

Divide each term in

2 y = x - 1

$2 y = x - 1$ by

2

$2$ .

\frac{2 y}{2} = \frac{x}{2} + \frac{- 1}{2}

$\frac{2 y}{2} = \frac{x}{2} + \frac{- 1}{2}$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of

2

$2$ .

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Step 3.3.2.1.1

Cancel the common factor.

\frac{2 y}{2} = \frac{x}{2} + \frac{- 1}{2}

$\frac{2 y}{2} = \frac{x}{2} + \frac{- 1}{2}$

Step 3.3.2.1.2

Divide

y

$y$ by

1

$1$ .

y = \frac{x}{2} + \frac{- 1}{2}

$y = \frac{x}{2} + \frac{- 1}{2}$

y = \frac{x}{2} + \frac{- 1}{2}

$y = \frac{x}{2} + \frac{- 1}{2}$

y = \frac{x}{2} + \frac{- 1}{2}

$y = \frac{x}{2} + \frac{- 1}{2}$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Move the negative in front of the fraction.

y = \frac{x}{2} - \frac{1}{2}

$y = \frac{x}{2} - \frac{1}{2}$

y = \frac{x}{2} - \frac{1}{2}

$y = \frac{x}{2} - \frac{1}{2}$

y = \frac{x}{2} - \frac{1}{2}

$y = \frac{x}{2} - \frac{1}{2}$

y = \frac{x}{2} - \frac{1}{2}

$y = \frac{x}{2} - \frac{1}{2}$

Step 4

Replace

y

$y$ with

f^{- 1} (x)

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

f^{- 1} (x) = \frac{x}{2} - \frac{1}{2}

$f^{- 1} (x) = \frac{x}{2} - \frac{1}{2}$

Step 5

Verify if

f^{- 1} (x) = \frac{x}{2} - \frac{1}{2}

$f^{- 1} (x) = \frac{x}{2} - \frac{1}{2}$ is the inverse of

f (x) = 2 x + 1

$f (x) = 2 x + 1$ .

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Step 5.1

To verify the inverse, check if

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

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

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

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

Step 5.2

Evaluate

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

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

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Step 5.2.1

Set up the composite result function.

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

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

Step 5.2.2

Evaluate

f^{- 1} (2 x + 1)

$f^{- 1} (2 x + 1)$ by substituting in the value of

f

$f$ into

f^{- 1}

$f^{- 1}$ .

f^{- 1} (2 x + 1) = \frac{2 x + 1}{2} - \frac{1}{2}

$f^{- 1} (2 x + 1) = \frac{2 x + 1}{2} - \frac{1}{2}$

Step 5.2.3

Combine the numerators over the common denominator.

f^{- 1} (2 x + 1) = \frac{2 x + 1 - 1}{2}

$f^{- 1} (2 x + 1) = \frac{2 x + 1 - 1}{2}$

Step 5.2.4

Combine the opposite terms in

2 x + 1 - 1

$2 x + 1 - 1$ .

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Step 5.2.4.1

Subtract

1

$1$ from

1

$1$ .

f^{- 1} (2 x + 1) = \frac{2 x + 0}{2}

$f^{- 1} (2 x + 1) = \frac{2 x + 0}{2}$

Step 5.2.4.2

Add

2 x

$2 x$ and

0

$0$ .

f^{- 1} (2 x + 1) = \frac{2 x}{2}

$f^{- 1} (2 x + 1) = \frac{2 x}{2}$

f^{- 1} (2 x + 1) = \frac{2 x}{2}

$f^{- 1} (2 x + 1) = \frac{2 x}{2}$

Step 5.2.5

Cancel the common factor of

2

$2$ .

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Step 5.2.5.1

Cancel the common factor.

f^{- 1} (2 x + 1) = \frac{2 x}{2}

$f^{- 1} (2 x + 1) = \frac{2 x}{2}$

Step 5.2.5.2

Divide

x

$x$ by

1

$1$ .

f^{- 1} (2 x + 1) = x

$f^{- 1} (2 x + 1) = x$

f^{- 1} (2 x + 1) = x

$f^{- 1} (2 x + 1) = x$

f^{- 1} (2 x + 1) = x

$f^{- 1} (2 x + 1) = x$

Step 5.3

Evaluate

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

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

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Step 5.3.1

Set up the composite result function.

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

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

Step 5.3.2

Evaluate

f (\frac{x}{2} - \frac{1}{2})

$f (\frac{x}{2} - \frac{1}{2})$ by substituting in the value of

f^{- 1}

$f^{- 1}$ into

f

$f$ .

f (\frac{x}{2} - \frac{1}{2}) = 2 (\frac{x}{2} - \frac{1}{2}) + 1

$f (\frac{x}{2} - \frac{1}{2}) = 2 (\frac{x}{2} - \frac{1}{2}) + 1$

Step 5.3.3

Simplify each term.

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Step 5.3.3.1

Apply the distributive property.

f (\frac{x}{2} - \frac{1}{2}) = 2 (\frac{x}{2}) + 2 (- \frac{1}{2}) + 1

$f (\frac{x}{2} - \frac{1}{2}) = 2 (\frac{x}{2}) + 2 (- \frac{1}{2}) + 1$

Step 5.3.3.2

Cancel the common factor of

2

$2$ .

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Step 5.3.3.2.1

Cancel the common factor.

f (\frac{x}{2} - \frac{1}{2}) = 2 (\frac{x}{2}) + 2 (- \frac{1}{2}) + 1

$f (\frac{x}{2} - \frac{1}{2}) = 2 (\frac{x}{2}) + 2 (- \frac{1}{2}) + 1$

Step 5.3.3.2.2

Rewrite the expression.

f (\frac{x}{2} - \frac{1}{2}) = x + 2 (- \frac{1}{2}) + 1

$f (\frac{x}{2} - \frac{1}{2}) = x + 2 (- \frac{1}{2}) + 1$

f (\frac{x}{2} - \frac{1}{2}) = x + 2 (- \frac{1}{2}) + 1

$f (\frac{x}{2} - \frac{1}{2}) = x + 2 (- \frac{1}{2}) + 1$

Step 5.3.3.3

Cancel the common factor of

2

$2$ .

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Step 5.3.3.3.1

Move the leading negative in

- \frac{1}{2}

$- \frac{1}{2}$ into the numerator.

f (\frac{x}{2} - \frac{1}{2}) = x + 2 (\frac{- 1}{2}) + 1

$f (\frac{x}{2} - \frac{1}{2}) = x + 2 (\frac{- 1}{2}) + 1$

Step 5.3.3.3.2

Cancel the common factor.

f (\frac{x}{2} - \frac{1}{2}) = x + 2 (\frac{- 1}{2}) + 1

$f (\frac{x}{2} - \frac{1}{2}) = x + 2 (\frac{- 1}{2}) + 1$

Step 5.3.3.3.3

Rewrite the expression.

f (\frac{x}{2} - \frac{1}{2}) = x - 1 + 1

$f (\frac{x}{2} - \frac{1}{2}) = x - 1 + 1$

f (\frac{x}{2} - \frac{1}{2}) = x - 1 + 1

$f (\frac{x}{2} - \frac{1}{2}) = x - 1 + 1$

f (\frac{x}{2} - \frac{1}{2}) = x - 1 + 1

$f (\frac{x}{2} - \frac{1}{2}) = x - 1 + 1$

Step 5.3.4

Combine the opposite terms in

x - 1 + 1

$x - 1 + 1$ .

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Step 5.3.4.1

Add

- 1

$- 1$ and

1

$1$ .

f (\frac{x}{2} - \frac{1}{2}) = x + 0

$f (\frac{x}{2} - \frac{1}{2}) = x + 0$

Step 5.3.4.2

Add

x

$x$ and

0

$0$ .

f (\frac{x}{2} - \frac{1}{2}) = x

$f (\frac{x}{2} - \frac{1}{2}) = x$

f (\frac{x}{2} - \frac{1}{2}) = x

$f (\frac{x}{2} - \frac{1}{2}) = x$

f (\frac{x}{2} - \frac{1}{2}) = x

$f (\frac{x}{2} - \frac{1}{2}) = x$

Step 5.4

Since

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

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

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

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

f^{- 1} (x) = \frac{x}{2} - \frac{1}{2}

$f^{- 1} (x) = \frac{x}{2} - \frac{1}{2}$ is the inverse of

f (x) = 2 x + 1

$f (x) = 2 x + 1$ .

f^{- 1} (x) = \frac{x}{2} - \frac{1}{2}

$f^{- 1} (x) = \frac{x}{2} - \frac{1}{2}$

f^{- 1} (x) = \frac{x}{2} - \frac{1}{2}

$f^{- 1} (x) = \frac{x}{2} - \frac{1}{2}$