Algebra Examples

y = - x^{2} - 3

$y = - x^{2} - 3$

Step 1

Interchange the variables.

x = - y^{2} - 3

$x = - y^{2} - 3$

Step 2

Solve for

y

$y$ .

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Step 2.1

Rewrite the equation as

- y^{2} - 3 = x

$- y^{2} - 3 = x$ .

- y^{2} - 3 = x

$- y^{2} - 3 = x$

Step 2.2

Add

3

$3$ to both sides of the equation.

- y^{2} = x + 3

$- y^{2} = x + 3$

Step 2.3

Divide each term in

- y^{2} = x + 3

$- y^{2} = x + 3$ by

- 1

$- 1$ and simplify.

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Step 2.3.1

Divide each term in

- y^{2} = x + 3

$- y^{2} = x + 3$ by

- 1

$- 1$ .

\frac{- y^{2}}{- 1} = \frac{x}{- 1} + \frac{3}{- 1}

$\frac{- y^{2}}{- 1} = \frac{x}{- 1} + \frac{3}{- 1}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Dividing two negative values results in a positive value.

\frac{y^{2}}{1} = \frac{x}{- 1} + \frac{3}{- 1}

$\frac{y^{2}}{1} = \frac{x}{- 1} + \frac{3}{- 1}$

Step 2.3.2.2

Divide

y^{2}

$y^{2}$ by

1

$1$ .

y^{2} = \frac{x}{- 1} + \frac{3}{- 1}

$y^{2} = \frac{x}{- 1} + \frac{3}{- 1}$

y^{2} = \frac{x}{- 1} + \frac{3}{- 1}

$y^{2} = \frac{x}{- 1} + \frac{3}{- 1}$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Simplify each term.

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Step 2.3.3.1.1

Move the negative one from the denominator of

\frac{x}{- 1}

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

y^{2} = - 1 \cdot x + \frac{3}{- 1}

$y^{2} = - 1 \cdot x + \frac{3}{- 1}$

Step 2.3.3.1.2

Rewrite

- 1 \cdot x

$- 1 \cdot x$ as

- x

$- x$ .

y^{2} = - x + \frac{3}{- 1}

$y^{2} = - x + \frac{3}{- 1}$

Step 2.3.3.1.3

Divide

3

$3$ by

- 1

$- 1$ .

y^{2} = - x - 3

$y^{2} = - x - 3$

y^{2} = - x - 3

$y^{2} = - x - 3$

y^{2} = - x - 3

$y^{2} = - x - 3$

y^{2} = - x - 3

$y^{2} = - x - 3$

Step 2.4

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

y = \pm \sqrt{- x - 3}

$y = \pm \sqrt{- x - 3}$

Step 2.5

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

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Step 2.5.1

First, use the positive value of the

\pm

$\pm$ to find the first solution.

y = \sqrt{- x - 3}

$y = \sqrt{- x - 3}$

Step 2.5.2

Next, use the negative value of the

\pm

$\pm$ to find the second solution.

y = - \sqrt{- x - 3}

$y = - \sqrt{- x - 3}$

Step 2.5.3

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

y = \sqrt{- x - 3}

$y = \sqrt{- x - 3}$

y = - \sqrt{- x - 3}

$y = - \sqrt{- x - 3}$

y = \sqrt{- x - 3}

$y = \sqrt{- x - 3}$

y = - \sqrt{- x - 3}

$y = - \sqrt{- x - 3}$

y = \sqrt{- x - 3}

$y = \sqrt{- x - 3}$

y = - \sqrt{- x - 3}

$y = - \sqrt{- x - 3}$

Step 3

Replace

y

$y$ with

f^{- 1} (x)

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

f^{- 1} (x) = \sqrt{- x - 3}, - \sqrt{- x - 3}

$f^{- 1} (x) = \sqrt{- x - 3}, - \sqrt{- x - 3}$

Step 4

Verify if

f^{- 1} (x) = \sqrt{- x - 3}, - \sqrt{- x - 3}

$f^{- 1} (x) = \sqrt{- x - 3}, - \sqrt{- x - 3}$ is the inverse of

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

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

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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} - 3

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

f^{- 1} (x) = \sqrt{- x - 3}, - \sqrt{- x - 3}

$f^{- 1} (x) = \sqrt{- x - 3}, - \sqrt{- x - 3}$ and compare them.

Step 4.2

Find the range of

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

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

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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:

(- \infty, - 3]

$(- \infty, - 3]$

(- \infty, - 3]

$(- \infty, - 3]$

Step 4.3

Find the domain of

\sqrt{- x - 3}

$\sqrt{- x - 3}$ .

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Step 4.3.1

Set the radicand in

\sqrt{- x - 3}

$\sqrt{- x - 3}$ greater than or equal to

0

$0$ to find where the expression is defined.

- x - 3 \geq 0

$- x - 3 \geq 0$

Step 4.3.2

Solve for

x

$x$ .

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Step 4.3.2.1

Add

3

$3$ to both sides of the inequality.

- x \geq 3

$- x \geq 3$

Step 4.3.2.2

Divide each term in

- x \geq 3

$- x \geq 3$ by

- 1

$- 1$ and simplify.

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Step 4.3.2.2.1

Divide each term in

- x \geq 3

$- x \geq 3$ by

- 1

$- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

\frac{- x}{- 1} \leq \frac{3}{- 1}

$\frac{- x}{- 1} \leq \frac{3}{- 1}$

Step 4.3.2.2.2

Simplify the left side.

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Step 4.3.2.2.2.1

Dividing two negative values results in a positive value.

\frac{x}{1} \leq \frac{3}{- 1}

$\frac{x}{1} \leq \frac{3}{- 1}$

Step 4.3.2.2.2.2

Divide

x

$x$ by

1

$1$ .

x \leq \frac{3}{- 1}

$x \leq \frac{3}{- 1}$

x \leq \frac{3}{- 1}

$x \leq \frac{3}{- 1}$

Step 4.3.2.2.3

Simplify the right side.

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Step 4.3.2.2.3.1

Divide

3

$3$ by

- 1

$- 1$ .

x \leq - 3

$x \leq - 3$

x \leq - 3

$x \leq - 3$

x \leq - 3

$x \leq - 3$

x \leq - 3

$x \leq - 3$

Step 4.3.3

The domain is all values of

x

$x$ that make the expression defined.

(- \infty, - 3]

$(- \infty, - 3]$

(- \infty, - 3]

$(- \infty, - 3]$

Step 4.4

Find the domain of

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

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

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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 - 3}, - \sqrt{- x - 3}

$f^{- 1} (x) = \sqrt{- x - 3}, - \sqrt{- x - 3}$ is the range of

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

$f (x) = - x^{2} - 3$ and the range of

f^{- 1} (x) = \sqrt{- x - 3}, - \sqrt{- x - 3}

$f^{- 1} (x) = \sqrt{- x - 3}, - \sqrt{- x - 3}$ is the domain of

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

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

f^{- 1} (x) = \sqrt{- x - 3}, - \sqrt{- x - 3}

$f^{- 1} (x) = \sqrt{- x - 3}, - \sqrt{- x - 3}$ is the inverse of

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

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

f^{- 1} (x) = \sqrt{- x - 3}, - \sqrt{- x - 3}

$f^{- 1} (x) = \sqrt{- x - 3}, - \sqrt{- x - 3}$

f^{- 1} (x) = \sqrt{- x - 3}, - \sqrt{- x - 3}

$f^{- 1} (x) = \sqrt{- x - 3}, - \sqrt{- x - 3}$

Step 5