Problem 32 - College Algebra

Question

In Exercises \(21-32,\) evaluate each function at the given values of the independent variable and simplify. $$f(x)=\frac{|x+3|}{x+3}$$ \({ a. } f(5) { b. } f(-5) { c. } f(-9-x)\)

Step-by-Step Solution

Answer

\({ a. } f(5) = 1, { b. } f(-5) = -1, { c. } f(-9-x) = 1 \) for \(x \leq -6\) and \(f(-9-x) = -1 \) for \(x > -6\).

Step 1: Evaluate the function at x = 5

Firstly, replace the variable \(x\) in the function with the value 5. This results in \(f(5) = \frac{|5 + 3|}{5 + 3}\). Calculate the numerator and then the denominator. The result is \(f(5)=\frac{8}{8}\). Hence, \(f(5)\) simplifies to 1.

Step 2: Evaluate the function at x = -5

Now, replace the variable \(x\) in the function with the value -5. This results in \(f(-5) = \frac{|-5 + 3|}{-5 + 3}\). After calculating inside the absolute value we get \(f(-5) = \frac{|-2|}{-2}\). The absolute value of -2 is 2, and thus we get \(f(-5) = \frac{2}{-2}\). Hence, \(f(-5)\) simplifies to -1.

Step 3: Evaluate the function at x = -9 - x

Finally, replace the variable \(x\) in the function with the value -9 - x. This results in \(f(-9 - x) = \frac{|-9 - x + 3|}{-9 - x + 3}\). This simplifies to \(f(-9 - x) = \frac{|-6 - x|}{-6 - x}\). Due to the absolute value, we have two possible results here. If \(-6 - x \geq 0\), then \(f(-9 - x) = 1\), and if \(-6 - x < 0\), then \(f(-9 - x) = -1\). Thus, the result depends on the specific value of \(x\).

Key Concepts

Understanding Absolute Value

The concept of absolute value helps us measure how far a number is from zero, regardless of its direction on the number line. Absolute value is always non-negative. For example, the absolute value of both 3 and -3 is 3, written as \(|3| = 3 = |-3|\). This is because distance is always positive.

In algebraic expressions, absolute values mean you take the absolute value of what is inside. For example, \(|x + 3|\) means you take the value of \(x+3\) and make it positive if it's not already.

• If \(x + 3\) is positive or zero, \(|x + 3| = x + 3\).
• If \(x + 3\) is negative, \(|x + 3| = -(x + 3)\).

Understanding this concept is essential when simplifying expressions that include absolute values. This is because we need to consider different scenarios for positive and negative inputs.

Exploring Piecewise Functions

Piecewise functions are special because they are defined by different expressions depending on the value of the input variable. This is exactly what we see with the function \(f(x)=\frac{|x+3|}{x+3}\).

The expression inside the absolute value, \(x+3\), dictates which piece of the function you're using:

• If \(x + 3 > 0\), \(f(x) = 1\) because the absolute value just returns \(x+3\) and it cancels out the denominator.
• If \(x + 3 < 0\), \(f(x) = -1\) because the absolute value flips \(x+3\) to \(-(x+3)\), resulting in \(-1\).

When \(x + 3 = 0\), the function is undefined because division by zero is not possible.

Piecewise functions like this one are handy when different conditions lead to different outcomes.

Simplification in Algebra

Simplification in algebra involves making an expression easier to read or solve, often by reducing it to the simplest form. This process includes:

• Combining like terms.
• Canceling common factors in fractions.
• Applying mathematical operations like addition, subtraction, multiplication, or division.

In the exercise, notice how the function \(f(x)=\frac{|x+3|}{x+3}\) simplifies in different contexts to either \(1\) or \(-1\).

To determine the simplest form, follow the rules of absolute values. Next, consider the expression without the absolute value to check if it can further reduce.

Simplification is crucial when solving equations, as it helps identify and eliminate unneeded complexity, making it easier to interpret and solve problems.