Algebra Examples

f (x) = 3 x - 2

$f (x) = 3 x - 2$

Step 1

Write

f (x) = 3 x - 2

$f (x) = 3 x - 2$ as an equation.

y = 3 x - 2

$y = 3 x - 2$

Step 2

Interchange the variables.

x = 3 y - 2

$x = 3 y - 2$

Step 3

Solve for

y

$y$ .

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Step 3.1

Rewrite the equation as

3 y - 2 = x

$3 y - 2 = x$ .

3 y - 2 = x

$3 y - 2 = x$

Step 3.2

Add

2

$2$ to both sides of the equation.

3 y = x + 2

$3 y = x + 2$

Step 3.3

Divide each term in

3 y = x + 2

$3 y = x + 2$ by

3

$3$ and simplify.

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Step 3.3.1

Divide each term in

3 y = x + 2

$3 y = x + 2$ by

3

$3$ .

\frac{3 y}{3} = \frac{x}{3} + \frac{2}{3}

$\frac{3 y}{3} = \frac{x}{3} + \frac{2}{3}$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of

3

$3$ .

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Step 3.3.2.1.1

Cancel the common factor.

\frac{3 y}{3} = \frac{x}{3} + \frac{2}{3}

$\frac{3 y}{3} = \frac{x}{3} + \frac{2}{3}$

Step 3.3.2.1.2

Divide

y

$y$ by

1

$1$ .

y = \frac{x}{3} + \frac{2}{3}

$y = \frac{x}{3} + \frac{2}{3}$

y = \frac{x}{3} + \frac{2}{3}

$y = \frac{x}{3} + \frac{2}{3}$

y = \frac{x}{3} + \frac{2}{3}

$y = \frac{x}{3} + \frac{2}{3}$

y = \frac{x}{3} + \frac{2}{3}

$y = \frac{x}{3} + \frac{2}{3}$

y = \frac{x}{3} + \frac{2}{3}

$y = \frac{x}{3} + \frac{2}{3}$

Step 4

Replace

y

$y$ with

f^{- 1} (x)

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

f^{- 1} (x) = \frac{x}{3} + \frac{2}{3}

$f^{- 1} (x) = \frac{x}{3} + \frac{2}{3}$

Step 5

Verify if

f^{- 1} (x) = \frac{x}{3} + \frac{2}{3}

$f^{- 1} (x) = \frac{x}{3} + \frac{2}{3}$ is the inverse of

f (x) = 3 x - 2

$f (x) = 3 x - 2$ .

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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} (3 x - 2)

$f^{- 1} (3 x - 2)$ by substituting in the value of

f

$f$ into

f^{- 1}

$f^{- 1}$ .

f^{- 1} (3 x - 2) = \frac{3 x - 2}{3} + \frac{2}{3}

$f^{- 1} (3 x - 2) = \frac{3 x - 2}{3} + \frac{2}{3}$

Step 5.2.3

Combine the numerators over the common denominator.

f^{- 1} (3 x - 2) = \frac{3 x - 2 + 2}{3}

$f^{- 1} (3 x - 2) = \frac{3 x - 2 + 2}{3}$

Step 5.2.4

Combine the opposite terms in

3 x - 2 + 2

$3 x - 2 + 2$ .

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Step 5.2.4.1

Add

- 2

$- 2$ and

2

$2$ .

f^{- 1} (3 x - 2) = \frac{3 x + 0}{3}

$f^{- 1} (3 x - 2) = \frac{3 x + 0}{3}$

Step 5.2.4.2

Add

3 x

$3 x$ and

0

$0$ .

f^{- 1} (3 x - 2) = \frac{3 x}{3}

$f^{- 1} (3 x - 2) = \frac{3 x}{3}$

f^{- 1} (3 x - 2) = \frac{3 x}{3}

$f^{- 1} (3 x - 2) = \frac{3 x}{3}$

Step 5.2.5

Cancel the common factor of

3

$3$ .

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Step 5.2.5.1

Cancel the common factor.

f^{- 1} (3 x - 2) = \frac{3 x}{3}

$f^{- 1} (3 x - 2) = \frac{3 x}{3}$

Step 5.2.5.2

Divide

x

$x$ by

1

$1$ .

f^{- 1} (3 x - 2) = x

$f^{- 1} (3 x - 2) = x$

f^{- 1} (3 x - 2) = x

$f^{- 1} (3 x - 2) = x$

f^{- 1} (3 x - 2) = x

$f^{- 1} (3 x - 2) = x$

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}{3} + \frac{2}{3})

$f (\frac{x}{3} + \frac{2}{3})$ by substituting in the value of

f^{- 1}

$f^{- 1}$ into

f

$f$ .

f (\frac{x}{3} + \frac{2}{3}) = 3 (\frac{x}{3} + \frac{2}{3}) - 2

$f (\frac{x}{3} + \frac{2}{3}) = 3 (\frac{x}{3} + \frac{2}{3}) - 2$

Step 5.3.3

Simplify each term.

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Step 5.3.3.1

Apply the distributive property.

f (\frac{x}{3} + \frac{2}{3}) = 3 (\frac{x}{3}) + 3 (\frac{2}{3}) - 2

$f (\frac{x}{3} + \frac{2}{3}) = 3 (\frac{x}{3}) + 3 (\frac{2}{3}) - 2$

Step 5.3.3.2

Cancel the common factor of

3

$3$ .

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Step 5.3.3.2.1

Cancel the common factor.

f (\frac{x}{3} + \frac{2}{3}) = 3 (\frac{x}{3}) + 3 (\frac{2}{3}) - 2

$f (\frac{x}{3} + \frac{2}{3}) = 3 (\frac{x}{3}) + 3 (\frac{2}{3}) - 2$

Step 5.3.3.2.2

Rewrite the expression.

f (\frac{x}{3} + \frac{2}{3}) = x + 3 (\frac{2}{3}) - 2

$f (\frac{x}{3} + \frac{2}{3}) = x + 3 (\frac{2}{3}) - 2$

f (\frac{x}{3} + \frac{2}{3}) = x + 3 (\frac{2}{3}) - 2

$f (\frac{x}{3} + \frac{2}{3}) = x + 3 (\frac{2}{3}) - 2$

Step 5.3.3.3

Cancel the common factor of

3

$3$ .

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Step 5.3.3.3.1

Cancel the common factor.

f (\frac{x}{3} + \frac{2}{3}) = x + 3 (\frac{2}{3}) - 2

$f (\frac{x}{3} + \frac{2}{3}) = x + 3 (\frac{2}{3}) - 2$

Step 5.3.3.3.2

Rewrite the expression.

f (\frac{x}{3} + \frac{2}{3}) = x + 2 - 2

$f (\frac{x}{3} + \frac{2}{3}) = x + 2 - 2$

f (\frac{x}{3} + \frac{2}{3}) = x + 2 - 2

$f (\frac{x}{3} + \frac{2}{3}) = x + 2 - 2$

f (\frac{x}{3} + \frac{2}{3}) = x + 2 - 2

$f (\frac{x}{3} + \frac{2}{3}) = x + 2 - 2$

Step 5.3.4

Combine the opposite terms in

x + 2 - 2

$x + 2 - 2$ .

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Step 5.3.4.1

Subtract

2

$2$ from

2

$2$ .

f (\frac{x}{3} + \frac{2}{3}) = x + 0

$f (\frac{x}{3} + \frac{2}{3}) = x + 0$

Step 5.3.4.2

Add

x

$x$ and

0

$0$ .

f (\frac{x}{3} + \frac{2}{3}) = x

$f (\frac{x}{3} + \frac{2}{3}) = x$

f (\frac{x}{3} + \frac{2}{3}) = x

$f (\frac{x}{3} + \frac{2}{3}) = x$

f (\frac{x}{3} + \frac{2}{3}) = x

$f (\frac{x}{3} + \frac{2}{3}) = x$

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}{3} + \frac{2}{3}

$f^{- 1} (x) = \frac{x}{3} + \frac{2}{3}$ is the inverse of

f (x) = 3 x - 2

$f (x) = 3 x - 2$ .

f^{- 1} (x) = \frac{x}{3} + \frac{2}{3}

$f^{- 1} (x) = \frac{x}{3} + \frac{2}{3}$

f^{- 1} (x) = \frac{x}{3} + \frac{2}{3}

$f^{- 1} (x) = \frac{x}{3} + \frac{2}{3}$

[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$