Problem

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

The problem is asking for two sets in mathematical analysis known as the domain and the range of the given function. The domain is the set of all possible input values (x-values) for which the function is defined, and the range is the set of all possible output values (y-values) that the function can produce. The function in question is \( f(x) = \frac{1}{1+\sqrt{x}} \). The task is to determine all the values of x that can be plugged into this function without causing any mathematical errors (like division by zero or taking the square root of a negative number), and then to find all the corresponding values of y that result from those x-values.

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

Answer

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Solution:

Step 1:

Ensure the expression under the square root, $x$, is non-negative to determine the valid input values.

$x \geq 0$

Step 2:

Identify values that make the denominator of $\frac{1}{1 + \sqrt{x}}$ zero to find where the function is not defined.

$1 + \sqrt{x} = 0$

Step 3:

Isolate $x$ to solve the equation.

Step 3.1:

Subtract $1$ from both sides to isolate the square root.

$\sqrt{x} = -1$

Step 3.2:

Square both sides to eliminate the square root.

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

Step 3.3:

Simplify both sides of the equation.

Step 3.3.1:

Express $\sqrt{x}$ as $x^{\frac{1}{2}}$ using the property $\sqrt[n]{a^x} = a^{\frac{x}{n}}$.

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

Step 3.3.2:

Simplify the left-hand side.

Step 3.3.2.1:

Apply the exponent rule $(a^m)^n = a^{mn}$.

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

Step 3.3.2.2:

Simplify by canceling out the common factors.

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

Step 3.3.2.3:

Rewrite the simplified expression.

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

Step 3.3.3:

Simplify the right-hand side.

$x = (-1)^2$

Step 3.4:

Discard any solutions that do not satisfy the original equation $1 + \sqrt{x} = 0$.

No solutions exist for this equation.

Step 4:

The domain consists of all $x$ values that keep the function defined.

Interval Notation: $[0, \infty)$ Set-Builder Notation: $\{x | x \geq 0\}$

Step 5:

Determine the range by considering all possible $y$ values output by the function.

Interval Notation: $(0, 1]$ Set-Builder Notation: $\{y | 0 < y \leq 1\}$

Step 6:

Combine the results to state the domain and range.

Domain: $[0, \infty)$, $\{x | x \geq 0\}$ Range: $(0, 1]$, $\{y | 0 < y \leq 1\}$

Knowledge Notes:

The domain of a function is the set of all possible input values (usually $x$) for which the function is defined. To find the domain of a function involving a square root, we must ensure that the expression under the square root (the radicand) is non-negative, as the square root of a negative number is not a real number.

The range of a function is the set of all possible output values (usually $y$) that the function can produce. For rational functions with a variable in the denominator, we must ensure that the denominator never equals zero, as division by zero is undefined.

Interval notation is a way of writing subsets of the real number line. An interval is written with a pair of numbers indicating the endpoints; brackets [ ] are used to indicate that an endpoint is included in the interval (closed interval), while parentheses ( ) indicate that an endpoint is not included (open interval).

Set-builder notation is another way to describe a set, defining the properties that its members must satisfy. It typically includes a variable, a vertical bar or colon, and a statement about the variable's properties.

When dealing with functions that include square roots or rational expressions, it is important to consider both the domain and range to fully understand the behavior of the function.

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