Find the Domain of the Quotient of the Two Functions f(x)=1/(x^2) , g(x) = square root of 2+x

$f (x) = \frac{1}{x^{2}}$ , $g (x) = \sqrt{2 + x}$

Find the quotient of the functions.

Replace the function designators with the actual functions in $\frac{f (x)}{g (x)}$ .

$\frac{\frac{1}{x^{2}}}{\sqrt{2 + x}}$

Simplify.

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{x^{2}} \cdot \frac{1}{\sqrt{2 + x}}$

Combine.

$\frac{1 \cdot 1}{x^{2} \sqrt{2 + x}}$

Multiply $1$ by $1$ .

$\frac{1}{x^{2} \sqrt{2 + x}}$

Multiply $\frac{1}{x^{2} \sqrt{2 + x}}$ by $\frac{\sqrt{2 + x}}{\sqrt{2 + x}}$ .

$\frac{1}{x^{2} \sqrt{2 + x}} \cdot \frac{\sqrt{2 + x}}{\sqrt{2 + x}}$

Combine and simplify the denominator.

Multiply $\frac{1}{x^{2} \sqrt{2 + x}}$ by $\frac{\sqrt{2 + x}}{\sqrt{2 + x}}$ .

$\frac{\sqrt{2 + x}}{x^{2} \sqrt{2 + x} \sqrt{2 + x}}$

Move $\sqrt{2 + x}$ .

$\frac{\sqrt{2 + x}}{x^{2} (\sqrt{2 + x} \sqrt{2 + x})}$

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

$\frac{\sqrt{2 + x}}{x^{2} ({\sqrt{2 + x}}^{1} \sqrt{2 + x})}$

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

$\frac{\sqrt{2 + x}}{x^{2} ({\sqrt{2 + x}}^{1} {\sqrt{2 + x}}^{1})}$

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

$\frac{\sqrt{2 + x}}{x^{2} {\sqrt{2 + x}}^{1 + 1}}$

Add $1$ and $1$ .

$\frac{\sqrt{2 + x}}{x^{2} {\sqrt{2 + x}}^{2}}$

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

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

$\frac{\sqrt{2 + x}}{x^{2} {({(2 + x)}^{\frac{1}{2}})}^{2}}$

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

$\frac{\sqrt{2 + x}}{x^{2} {(2 + x)}^{\frac{1}{2} \cdot 2}}$

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

$\frac{\sqrt{2 + x}}{x^{2} {(2 + x)}^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{\sqrt{2 + x}}{x^{2} {(2 + x)}^{\frac{2}{2}}}$

Rewrite the expression.

$\frac{\sqrt{2 + x}}{x^{2} {(2 + x)}^{1}}$

Simplify.

$\frac{\sqrt{2 + x}}{x^{2} (2 + x)}$

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

$2 + x \geq 0$

Subtract $2$ from both sides of the inequality.

$x \geq - 2$

Set the denominator in $\frac{\sqrt{2 + x}}{x^{2} (2 + x)}$ equal to $0$ to find where the expression is undefined.

$x^{2} (2 + x) = 0$

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 $0$ .

$x^{2} = 0$

$2 + x = 0$

Set $x^{2}$ equal to $0$ and solve for $x$ .

Set $x^{2}$ equal to $0$ .

$x^{2} = 0$

Solve $x^{2} = 0$ for $x$ .

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

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

Set $2 + x$ equal to $0$ and solve for $x$ .

Set $2 + x$ equal to $0$ .

$2 + x = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

The final solution is all the values that make $x^{2} (2 + x) = 0$ true.

$x = 0, - 2$

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

Interval Notation:

$(- 2, 0) \cup (0, \infty)$

Set-Builder Notation:

${x | x > - 2, x \neq 0}$

Find the Domain of the Quotient of the Two Functions f(x)=1/(x^2) , g(x) = square root of 2+x

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