Algebra Examples

y = x^{2} - 12

$y = x^{2} - 12$

Step 1

Interchange the variables.

x = y^{2} - 12

$x = y^{2} - 12$

Step 2

Solve for

y

$y$ .

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Step 2.1

Rewrite the equation as

y^{2} - 12 = x

$y^{2} - 12 = x$ .

y^{2} - 12 = x

$y^{2} - 12 = x$

Step 2.2

Add

12

$12$ to both sides of the equation.

y^{2} = x + 12

$y^{2} = x + 12$

Step 2.3

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

y = \pm \sqrt{x + 12}

$y = \pm \sqrt{x + 12}$

Step 2.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 2.4.1

First, use the positive value of the

\pm

$\pm$ to find the first solution.

y = \sqrt{x + 12}

$y = \sqrt{x + 12}$

Step 2.4.2

Next, use the negative value of the

\pm

$\pm$ to find the second solution.

y = - \sqrt{x + 12}

$y = - \sqrt{x + 12}$

Step 2.4.3

The complete solution is the result of both the positive and negative portions of the solution.

y = \sqrt{x + 12}

$y = \sqrt{x + 12}$

y = - \sqrt{x + 12}

$y = - \sqrt{x + 12}$

y = \sqrt{x + 12}

$y = \sqrt{x + 12}$

y = - \sqrt{x + 12}

$y = - \sqrt{x + 12}$

y = \sqrt{x + 12}

$y = \sqrt{x + 12}$

y = - \sqrt{x + 12}

$y = - \sqrt{x + 12}$

Step 3

Replace

y

$y$ with

f^{- 1} (x)

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

f^{- 1} (x) = \sqrt{x + 12}, - \sqrt{x + 12}

$f^{- 1} (x) = \sqrt{x + 12}, - \sqrt{x + 12}$

Step 4

Verify if

f^{- 1} (x) = \sqrt{x + 12}, - \sqrt{x + 12}

$f^{- 1} (x) = \sqrt{x + 12}, - \sqrt{x + 12}$ is the inverse of

f (x) = x^{2} - 12

$f (x) = x^{2} - 12$ .

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Step 4.1

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of

f (x) = x^{2} - 12

$f (x) = x^{2} - 12$ and

f^{- 1} (x) = \sqrt{x + 12}, - \sqrt{x + 12}

$f^{- 1} (x) = \sqrt{x + 12}, - \sqrt{x + 12}$ and compare them.

Step 4.2

Find the range of

f (x) = x^{2} - 12

$f (x) = x^{2} - 12$ .

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Step 4.2.1

The range is the set of all valid

y

$y$ values. Use the graph to find the range.

Interval Notation:

[- 12, \infty)

$[- 12, \infty)$

[- 12, \infty)

$[- 12, \infty)$

Step 4.3

Find the domain of

\sqrt{x + 12}

$\sqrt{x + 12}$ .

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Step 4.3.1

Set the radicand in

\sqrt{x + 12}

$\sqrt{x + 12}$ greater than or equal to

0

$0$ to find where the expression is defined.

x + 12 \geq 0

$x + 12 \geq 0$

Step 4.3.2

Subtract

12

$12$ from both sides of the inequality.

x \geq - 12

$x \geq - 12$

Step 4.3.3

The domain is all values of

x

$x$ that make the expression defined.

[- 12, \infty)

$[- 12, \infty)$

[- 12, \infty)

$[- 12, \infty)$

Step 4.4

Find the domain of

f (x) = x^{2} - 12

$f (x) = x^{2} - 12$ .

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Step 4.4.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

(- \infty, \infty)

$(- \infty, \infty)$

(- \infty, \infty)

$(- \infty, \infty)$

Step 4.5

Since the domain of

f^{- 1} (x) = \sqrt{x + 12}, - \sqrt{x + 12}

$f^{- 1} (x) = \sqrt{x + 12}, - \sqrt{x + 12}$ is the range of

f (x) = x^{2} - 12

$f (x) = x^{2} - 12$ and the range of

f^{- 1} (x) = \sqrt{x + 12}, - \sqrt{x + 12}

$f^{- 1} (x) = \sqrt{x + 12}, - \sqrt{x + 12}$ is the domain of

f (x) = x^{2} - 12

$f (x) = x^{2} - 12$ , then

f^{- 1} (x) = \sqrt{x + 12}, - \sqrt{x + 12}

$f^{- 1} (x) = \sqrt{x + 12}, - \sqrt{x + 12}$ is the inverse of

f (x) = x^{2} - 12

$f (x) = x^{2} - 12$ .

f^{- 1} (x) = \sqrt{x + 12}, - \sqrt{x + 12}

$f^{- 1} (x) = \sqrt{x + 12}, - \sqrt{x + 12}$

f^{- 1} (x) = \sqrt{x + 12}, - \sqrt{x + 12}

$f^{- 1} (x) = \sqrt{x + 12}, - \sqrt{x + 12}$

Step 5