Find the Domain and Range x=2y^2

$x = 2 y^{2}$

Step 1

Rewrite the equation as $2 y^{2} = x$ .

$2 y^{2} = x$

Step 2

Divide each term in $2 y^{2} = x$ by $2$ and simplify.

Divide each term in $2 y^{2} = x$ by $2$ .

$\frac{2 y^{2}}{2} = \frac{x}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y^{2}}{2} = \frac{x}{2}$

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

$y^{2} = \frac{x}{2}$

Step 3

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

$y = \pm \sqrt{\frac{x}{2}}$

Step 4

Simplify $\pm \sqrt{\frac{x}{2}}$ .

Rewrite $\sqrt{\frac{x}{2}}$ as $\frac{\sqrt{x}}{\sqrt{2}}$ .

$y = \pm \frac{\sqrt{x}}{\sqrt{2}}$

Multiply $\frac{\sqrt{x}}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$y = \pm \frac{\sqrt{x}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$

Combine and simplify the denominator.

Multiply $\frac{\sqrt{x}}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$y = \pm \frac{\sqrt{x} \sqrt{2}}{\sqrt{2} \sqrt{2}}$

Raise $\sqrt{2}$ to the power of $1$ .

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

Raise $\sqrt{2}$ to the power of $1$ .

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

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

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

Add $1$ and $1$ .

$y = \pm \frac{\sqrt{x} \sqrt{2}}{{\sqrt{2}}^{2}}$

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

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

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

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

$y = \pm \frac{\sqrt{x} \sqrt{2}}{2^{\frac{1}{2} \cdot 2}}$

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

$y = \pm \frac{\sqrt{x} \sqrt{2}}{2^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$y = \pm \frac{\sqrt{x} \sqrt{2}}{2^{\frac{2}{2}}}$

Rewrite the expression.

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

Evaluate the exponent.

$y = \pm \frac{\sqrt{x} \sqrt{2}}{2}$

Combine using the product rule for radicals.

$y = \pm \frac{\sqrt{x \cdot 2}}{2}$

Reorder factors in $\pm \frac{\sqrt{x \cdot 2}}{2}$ .

$y = \pm \frac{\sqrt{2 x}}{2}$

Step 5

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.

$y = \frac{\sqrt{2 x}}{2}$

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

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

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

$y = \frac{\sqrt{2 x}}{2}$

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

$y = \frac{\sqrt{2 x}}{2}$

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

Step 6

Set the radicand in $\sqrt{2 x}$ greater than or equal to $0$ to find where the expression is defined.

$2 x \geq 0$

Step 7

Divide each term in $2 x \geq 0$ by $2$ and simplify.

Divide each term in $2 x \geq 0$ by $2$ .

$\frac{2 x}{2} \geq \frac{0}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} \geq \frac{0}{2}$

Divide $x$ by $1$ .

$x \geq \frac{0}{2}$

Simplify the right side.

Divide $0$ by $2$ .

$x \geq 0$

Step 8

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

Interval Notation:

$[0, \infty)$

Set-Builder Notation:

${x | x \geq 0}$

Step 9

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

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${y | y \in ℝ}$

Step 10

Determine the domain and range.

Domain: $[0, \infty), {x | x \geq 0}$

Range: $(- \infty, \infty), {y | y \in ℝ}$

Step 11

Find the Domain and Range x=2y^2

Related Questions

Support LCMGCF.com by disabling your adblocker.