Problem 37 - Algebra Form and Function

Question

For the functions, (a) List the algebraic operations in order of evaluation. What restrictions does each operation place on the domain of the function? (b) Give the function's domain. $$ y=4-(x-3)^{1 / 2} $$

Step-by-Step Solution

Answer

List the algebraic operations in order of evaluation and the restrictions each operation places on the domain of the function: 1. $$x - 3$$ - No restrictions. 2. $$(x-3)^{1/2}$$ - $$x-3$$ must be non-negative (i.e., $$x-3 \ge 0$$). 3. $$4 - (x-3)^{1/2}$$ - No restrictions. Function's domain: $$x\ge 3$$.

Step 1: Identify Algebraic Operations

The given function is $$y = 4 - (x-3)^{1/2}$$. There are 3 algebraic operations happening in order: 1. $$x - 3$$ 2. $$(x-3)^{1/2}$$ 3. $$4 - (x-3)^{1/2}$$ Next, let's examine the restrictions each operation places on the domain.

Step 2: Determine Restrictions on Domain

Going through each operation again: 1. $$x - 3$$ - This operation has no restriction on the domain. 2. $$(x-3)^{1/2}$$ - We are taking the square root of the expression $$x-3$$. Recall that we cannot take the square root of a negative number, so $$x-3$$ must be non-negative (i.e., $$x-3 \ge 0$$). 3. $$4 - (x-3)^{1/2}$$ - After taking the square root, we are subtracting the result from 4, which places no restrictions on the domain. From these operations, the only restriction on the domain comes from operation 2. Therefore, the domain of the function has to satisfy $$x-3\ge 0$$.

Step 3: Domain of the Function

With the restrictions identified in the previous step, let's now find the domain of the function: 1. From operation 2, we know $$x-3\ge 0$$. 2. Solving for $$x$$, we get $$x\ge3$$. Thus, the domain of the function $$y = 4 - (x-3)^{1/2}$$ is $$x\ge3$$.

Key Concepts

Algebraic Operations

When we deal with functions, understanding the sequence of algebraic operations is key. For our function \( y = 4 - (x-3)^{1/2} \), there are three main steps.

• First, we perform the operation \( x-3 \). This is a simple subtraction and has no specific restrictions on the values \( x \) can take. It can handle any real number.
• Next comes the operation \( (x-3)^{1/2} \), which involves taking the square root of the result from the previous operation. This is a bit more complex since it adds a restriction on the domain.
• Lastly, we have the subtraction \( 4 - (x-3)^{1/2} \). This combines the result with 4 and does not impose any further restrictions.

Breaking down functions into these algebraic operations helps us better understand any domain limitations.

Domain Restrictions

Every algebraic operation comes with its own set of rules, and some affect the domain of the function. The function's domain is basically all the possible values of \( x \) that make the function work without any hiccups.

• For our operation \( x - 3 \), no restriction means \( x \) can be any number.
• However, the square root operation \( (x-3)^{1/2} \) requires that \( x-3 \) is not negative. This sets the condition \( x-3 \ge 0 \). Solving this inequality gives us \( x \ge 3 \).

Thus, the domain of our function is limited to \( x \ge 3 \), as any less would make the expression inside the square root negative, which is not defined among real numbers.

Square Root Function

The square root function is one of those special functions in math with its own set of rules. It only works where it can produce real number results.

• The operation \( (x-3)^{1/2} \) means we take the square root of \( x-3 \). In the real number system, you cannot take the square root of a negative number.
• This places an important limitation or restriction: \( x-3 \) must be zero or positive. Therefore, \( x \) must be at least 3, as identified earlier.

Remember, the square root is like a safety gate that allows only certain values to pass through, making it essential to respect its restrictions to find the function's valid domain.