Find the Domain and Range xy^2-x+3y^2+1=0

$x y^{2} - x + 3 y^{2} + 1 = 0$

Step 1

Move all terms not containing $y$ to the right side of the equation.

Add $x$ to both sides of the equation.

$x y^{2} + 3 y^{2} + 1 = x$

Subtract $1$ from both sides of the equation.

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

Step 2

Factor $y^{2}$ out of $x y^{2} + 3 y^{2}$ .

Factor $y^{2}$ out of $x y^{2}$ .

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

Factor $y^{2}$ out of $3 y^{2}$ .

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

Factor $y^{2}$ out of $y^{2} x + y^{2} \cdot 3$ .

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

Step 3

Divide each term in $y^{2} (x + 3) = x - 1$ by $x + 3$ and simplify.

Divide each term in $y^{2} (x + 3) = x - 1$ by $x + 3$ .

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

Simplify the left side.

Cancel the common factor of $x + 3$ .

Cancel the common factor.

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

Divide $y^{2}$ by $1$ .

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

Simplify the right side.

Combine the numerators over the common denominator.

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

Step 4

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

$y = \pm \sqrt{\frac{x - 1}{x + 3}}$

Step 5

Simplify $\pm \sqrt{\frac{x - 1}{x + 3}}$ .

Rewrite $\sqrt{\frac{x - 1}{x + 3}}$ as $\frac{\sqrt{x - 1}}{\sqrt{x + 3}}$ .

$y = \pm \frac{\sqrt{x - 1}}{\sqrt{x + 3}}$

Multiply $\frac{\sqrt{x - 1}}{\sqrt{x + 3}}$ by $\frac{\sqrt{x + 3}}{\sqrt{x + 3}}$ .

$y = \pm \frac{\sqrt{x - 1}}{\sqrt{x + 3}} \cdot \frac{\sqrt{x + 3}}{\sqrt{x + 3}}$

Combine and simplify the denominator.

Multiply $\frac{\sqrt{x - 1}}{\sqrt{x + 3}}$ by $\frac{\sqrt{x + 3}}{\sqrt{x + 3}}$ .

$y = \pm \frac{\sqrt{x - 1} \sqrt{x + 3}}{\sqrt{x + 3} \sqrt{x + 3}}$

Raise $\sqrt{x + 3}$ to the power of $1$ .

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

Raise $\sqrt{x + 3}$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $1$ .

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

Rewrite ${\sqrt{x + 3}}^{2}$ as $x + 3$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 3}$ as ${(x + 3)}^{\frac{1}{2}}$ .

$y = \pm \frac{\sqrt{x - 1} \sqrt{x + 3}}{{({(x + 3)}^{\frac{1}{2}})}^{2}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$y = \pm \frac{\sqrt{x - 1} \sqrt{x + 3}}{{(x + 3)}^{\frac{1}{2} \cdot 2}}$

Combine $\frac{1}{2}$ and $2$ .

$y = \pm \frac{\sqrt{x - 1} \sqrt{x + 3}}{{(x + 3)}^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$y = \pm \frac{\sqrt{x - 1} \sqrt{x + 3}}{{(x + 3)}^{\frac{2}{2}}}$

Rewrite the expression.

$y = \pm \frac{\sqrt{x - 1} \sqrt{x + 3}}{{(x + 3)}^{1}}$

Simplify.

$y = \pm \frac{\sqrt{x - 1} \sqrt{x + 3}}{x + 3}$

Combine using the product rule for radicals.

$y = \pm \frac{\sqrt{(x - 1) (x + 3)}}{x + 3}$

Step 6

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

First, use the positive value of the $\pm$ to find the first solution.

$y = \frac{\sqrt{(x - 1) (x + 3)}}{x + 3}$

Next, use the negative value of the $\pm$ to find the second solution.

$y = - \frac{\sqrt{(x - 1) (x + 3)}}{x + 3}$

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

$y = \frac{\sqrt{(x - 1) (x + 3)}}{x + 3}$

$y = - \frac{\sqrt{(x - 1) (x + 3)}}{x + 3}$

$y = \frac{\sqrt{(x - 1) (x + 3)}}{x + 3}$

$y = - \frac{\sqrt{(x - 1) (x + 3)}}{x + 3}$

Step 7

Set the radicand in $\sqrt{(x - 1) (x + 3)}$ greater than or equal to $0$ to find where the expression is defined.

$(x - 1) (x + 3) \geq 0$

Step 8

Solve for $x$ .

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 1 = 0$

$x + 3 = 0$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $(x - 1) (x + 3) \geq 0$ true.

$x = 1, - 3$

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 1$

$x > 1$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < - 3$ to see if it makes the inequality true.

Choose a value on the interval $x < - 3$ and see if this value makes the original inequality true.

$x = - 6$

Replace $x$ with $- 6$ in the original inequality.

$((- 6) - 1) ((- 6) + 3) \geq 0$

The left side $21$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 3 < x < 1$ to see if it makes the inequality true.

Choose a value on the interval $- 3 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$((0) - 1) ((0) + 3) \geq 0$

The left side $- 3$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 1$ to see if it makes the inequality true.

Choose a value on the interval $x > 1$ and see if this value makes the original inequality true.

$x = 4$

Replace $x$ with $4$ in the original inequality.

$((4) - 1) ((4) + 3) \geq 0$

The left side $21$ is greater than the right side $0$ , which means that the given statement is always true.

True

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 3$ True

$- 3 < x < 1$ False

$x > 1$ True

$x < - 3$ True

$- 3 < x < 1$ False

$x > 1$ True

The solution consists of all of the true intervals.

$x \leq - 3$ or $x \geq 1$

Step 9

Set the denominator in $\frac{\sqrt{(x - 1) (x + 3)}}{x + 3}$ equal to $0$ to find where the expression is undefined.

$x + 3 = 0$

Step 10

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 11

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(- \infty, - 3) \cup [1, \infty)$

Set-Builder Notation:

${x | x < - 3, x \geq 1}$

Step 12

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$(- \infty, - 1) \cup (- 1, 1) \cup (1, \infty)$

Set-Builder Notation:

${y | y \neq 1, - 1}$

Step 13

Determine the domain and range.

Domain: $(- \infty, - 3) \cup [1, \infty), {x | x < - 3, x \geq 1}$

Range: $(- \infty, - 1) \cup (- 1, 1) \cup (1, \infty), {y | y \neq 1, - 1}$

Step 14

Find the Domain and Range xy^2-x+3y^2+1=0

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