Problem 38 - College Algebra Essentials

Question

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

The solutions are: a. \(f(5) = 1\) b. \(f(-5) = -1\) c. When \(x < -6\), \(f(x) = 1\) and when \(x > -6\), \(f(x) = -1\).

Step 1: Substitute given values into the function

For each subpart (a, b, c), substitute the given value into the function \(f(x)\) and simplify the equation accordingly.

Step 1a: Evaluate \(f(5)\)

For part 'a', \(f(5)\) substitute \(x = 5\) into the function to get \(f(5) = \frac{|5 + 3|}{5 + 3}\). Calculate to get \(f(5) = 1\).

Step 1b: Evaluate \(f(-5)\)

For part 'b', \(f(-5)\), substitute \(x = -5\) into the function to get \(f(-5) = \frac{|-5 + 3|}{-5 + 3}\). Calculate to get \(f(-5) = -1\).

Step 1c: Evaluate \(f(-9-x)\)

For part 'c', \(f(-9-x)\), substitute \(x = -9 - x\) into the function to get \(f(-9-x) = \frac{|-9 - x + 3|}{-9 - x + 3}\). Simplify the absolute value part \(-9 - x + 3\) to get \(-6 - x\). Consider the two cases for the absolute value: When \(-6 - x\) is positive, which occurs when \(x < -6\), the value inside the absolute value signs is positive, and we get \(f(x) = 1\). When \(-6 - x\) is negative, which occurs when \(x > -6\), the value inside the absolute value signs is negative, and we have to change the sign to get \(f(x) = -1\).

Key Concepts

Absolute Value

Understanding absolute value is essential when evaluating certain types of functions. The absolute value of a number is its distance from zero on the number line, regardless of the direction. It is always a positive number or zero. Absolute value is represented by two vertical bars, for example, \( |x+3| \). This means 'regardless of whether \( x+3 \) is positive or negative, take its positive value.'

In the given function \( f(x) = \frac{|x+3|}{x+3} \), the absolute value affects the numerator. Depending on the value of \( x \), the expression inside the absolute value can be negative or positive. The purpose of the absolute value is to ensure that the numerator is always non-negative.

• If \( x+3 > 0 \), then \( |x+3| = x+3 \).
• If \( x+3 < 0 \), then \( |x+3| = -(x+3) \).

Piecewise Functions

Piecewise functions are defined by multiple sub-functions, each applying to different parts of the function's domain. For example, \( f(x) = \frac{|x+3|}{x+3} \) can act like a piecewise function based on the value of \( x \). This means the function can take on different expressions depending on the input value.

This specific function has two cases because of the absolute value in the numerator:

• When \( x+3 > 0 \), the function simplifies to \( f(x) = 1 \), as the numerator and the denominator are equal and positive.
• When \( x+3 < 0 \), the function simplifies to \( f(x) = -1 \), because the numerator \( |x+3| \) is positive and the denominator \( x+3 \) is negative, making the overall fraction negative.

Understanding piecewise functions helps you anticipate and verify the behavior of complex functions.

Independent Variable Evaluation

Evaluating a function means substituting specific values for the independent variable, commonly \( x \), to find the corresponding function's output. In the exercise, you are given specific values to substitute into the function \( f(x) = \frac{|x+3|}{x+3} \).

Here's how you evaluate:

• Identify the given value for the independent variable \( x \).
• Substitute this value into the function, replacing all instances of \( x \) with the given number.
• Perform arithmetic operations as needed, following the order of operations: absolute values, multiplication and division, etc.

For example:

• To find \( f(5) \), replace \( x \) with \( 5 \), resulting in \( f(5) = \frac{|5+3|}{5+3} = \frac{8}{8} = 1 \).
• For \( f(-5) \), replace \( x \) with \( -5 \), giving \( f(-5) = \frac{|-5+3|}{-5+3} = \frac{2}{-2} = -1 \).
• The result may be affected by the structure of the function, like absolute values or potential piecewise definitions, as seen in \( f(-9-x) \).

Correctly following these steps ensures that you accurately compute the function for any given value of \( x \).