Problem 24 - College Algebra with Modeling and Visualization

Question

Complete the following. (a) Find \(f(x)\) for the indicated values of \(x\), if possible. (b) Find the domain of \(f\). $$ f(x)=\sqrt{1-x} \text { for } x=-2, a+2 $$

Step-by-Step Solution

Answer

f(-2) = \(\sqrt{3}\), f(a+2) is undefined for real numbers. Domain of f is \((-\infty, 1]\).

Step 1: Evaluate f(x) for x=-2

To find \( f(-2) \), substitute \(-2\) for \( x \) in the function \( f(x) = \sqrt{1-x} \):\[f(-2) = \sqrt{1 - (-2)} = \sqrt{1 + 2} = \sqrt{3}.\]Thus, \( f(-2) = \sqrt{3} \).

Step 2: Evaluate f(x) for x=a+2

To find \( f(a+2) \), substitute \( a+2 \) for \( x \) in the function \( f(x) = \sqrt{1-x} \):\[f(a+2) = \sqrt{1 - (a+2)} = \sqrt{1 - a - 2} = \sqrt{-a - 1}.\]This expression is only valid if \( -a - 1 \geq 0 \), which is not possible for non-complex numbers. Therefore, \( f(a+2) \) cannot be evaluated for real numbers.

Step 3: Determine the Domain of f(x)

To find the domain of \( f(x) = \sqrt{1-x} \), we need to ensure the expression under the square root is non-negative:\[1-x \geq 0.\]Rearrange to find \( x \leq 1 \). Thus, the domain of \( f \) is all \( x \) such that \( x \leq 1 \), or in interval notation, \((-\infty, 1]\).

Key Concepts

Domain of a Function

Understanding the domain of a function is crucial because it tells us the set of all possible inputs (or x-values) for which the function is defined. For the function \( f(x) = \sqrt{1-x} \), finding the domain involves ensuring that the expression inside the square root is non-negative. This is important because the square root of a negative number is not real.
To determine this, we analyze the inequality \( 1-x \geq 0 \). Solving for \( x \), we rearrange it to get \( x \leq 1 \). This means that our function \( f(x) \) is only defined for values of \( x \) that are less than or equal to 1, as those values make the expression under the square root valid and non-negative. Therefore, the domain of \( f(x) \) in interval notation is \((-\infty, 1]\).
The domain is important not only for understanding functions but also for predicting their behavior and ensuring we're working with legitimate values.

Square Root Function

Square root functions are a type of function that includes a square root. For the function \( f(x) = \sqrt{1-x} \), the square root sign indicates that we should consider only the non-negative square root. Unlike functions with polynomials, where any real number can be an input, square root functions are restricted by their domains.
With square root functions, it's important to remember:

• The expression inside the square root must be non-negative, as square roots of negative numbers aren't real.
• The behavior of the function "ends" at the points where the inside of the square root equals zero, which defines the boundary of the domain.

In our specific function, \( f(x) = \sqrt{1-x} \), the curve of the graph will approach closer to zero as \( x \) approaches 1 from the negative side in the real number line, but it stops at \( x = 1 \). This is because square root functions are only real-valued when dealing with non-negative radicands.

Evaluation of Functions

Evaluating a function means finding the value of the function at a specific point or expression. For instance, with \( f(x) = \sqrt{1-x} \), evaluating \( f(-2) \) involves substituting \(-2\) into the function:
\[ f(-2) = \sqrt{1 - (-2)} = \sqrt{3}. \]
Here, the calculation was straightforward because the expression inside the square root, \(1 + 2\), was positive.
However, when evaluating \( f(a+2) \), we face a different situation:
\[ f(a+2) = \sqrt{1 - (a + 2)} = \sqrt{-a - 1}. \]
This expression cannot be evaluated in the real number system because the inside becomes negative. Hence, the function cannot be evaluated without specifying conditions where the inside is non-negative.

• This exercise highlights the need to check the validity of computations within the domain before carrying out evaluations.
• Always ensure compliance with the function's domain restrictions to avoid non-real solutions.