Find the Domain and Range f(x)=1/(1+ square root of x)

$f (x) = \frac{1}{1 + \sqrt{x}}$

Step 1

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

$x \geq 0$

Step 2

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

$1 + \sqrt{x} = 0$

Step 3

Solve for $x$ .

Subtract $1$ from both sides of the equation.

$\sqrt{x} = - 1$

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{x}}^{2} = {(- 1)}^{2}$

Simplify each side of the equation.

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

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

Simplify the left side.

Simplify ${(x^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${(x^{\frac{1}{2}})}^{2}$ .

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

$x^{\frac{1}{2} \cdot 2} = {(- 1)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$x^{\frac{1}{2} \cdot 2} = {(- 1)}^{2}$

Rewrite the expression.

$x^{1} = {(- 1)}^{2}$

Simplify.

$x = {(- 1)}^{2}$

Simplify the right side.

Raise $- 1$ to the power of $2$ .

$x = 1$

Exclude the solutions that do not make $1 + \sqrt{x} = 0$ true.

$No solution$

Step 4

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

Interval Notation:

$[0, \infty)$

Set-Builder Notation:

${x | x \geq 0}$

Step 5

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

Interval Notation:

$(0, 1]$

Set-Builder Notation:

${y | 0 < y \leq 1}$

Step 6

Determine the domain and range.

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

Range: $(0, 1], {y | 0 < y \leq 1}$

Step 7

Find the Domain and Range f(x)=1/(1+ square root of x)

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