Problem 26 - Algebra and Trigonometry

Question

Evaluate the function at the indicated values. $$ \begin{array}{l}{f(x)=\frac{|x|}{x}} \\ {f(-2), f(-1), f(0), f(5), f\left(x^{2}\right), f\left(\frac{1}{x}\right)}\end{array} $$

Step-by-Step Solution

Answer

\( f(-2) = -1 \), \( f(-1) = -1 \), \( f(0) \) is undefined, \( f(5) = 1 \), \( f(x^2) = 1 \) for \( x \neq 0 \), \( f(\frac{1}{x}) = 1 \) if \( x > 0 \), \( f(\frac{1}{x}) = -1 \) if \( x < 0 \).

Step 1: Understand the Function

The function given is \( f(x) = \frac{|x|}{x} \). This is a piecewise function. It can be broken down as follows:- When \( x > 0 \), \( f(x) = 1 \).- When \( x < 0 \), \( f(x) = -1 \).- When \( x = 0 \), the function is undefined because division by zero is not allowed.

Step 2: Evaluate \( f(-2) \)

Here, \( x = -2 \). Since \( x < 0 \), \( f(x) = -1 \). Therefore, \( f(-2) = -1 \).

Step 3: Evaluate \( f(-1) \)

Here, \( x = -1 \). As \( x < 0 \), \( f(x) = -1 \). Therefore, \( f(-1) = -1 \).

Step 4: Evaluate \( f(0) \)

Here, \( x = 0 \). The function is undefined at \( x = 0 \) due to division by zero. Thus, \( f(0) \) is undefined.

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

Here, \( x = 5 \). Since \( x > 0 \), \( f(x) = 1 \). Therefore, \( f(5) = 1 \).

Step 6: Evaluate \( f(x^2) \)

Here, the input is \( x^2 \). Since \( x^2 \geq 0 \) for all real \( x \), thus \( f(x^2) = 1 \) when \( x^2 > 0 \) and undefined when \( x = 0 \). Overall, \( f(x^2) = 1 \) for \( x eq 0 \) and undefined for \( x = 0 \).

Step 7: Evaluate \( f(\frac{1}{x}) \)

Here, the input is \( \frac{1}{x} \). If \( x > 0 \), then \( \frac{1}{x} > 0 \) and \( f(\frac{1}{x}) = 1 \). If \( x < 0 \), then \( \frac{1}{x} < 0 \) and \( f(\frac{1}{x}) = -1 \). If \( x = 0 \), \( \frac{1}{x} \) is undefined, so \( f(\frac{1}{x}) \) is also undefined.

Key Concepts

Absolute Value

The absolute value of a number is its distance from zero on a number line, regardless of direction. It's always non-negative, represented as \(|x|\). For example, \(|-3| = 3\) and \(|4| = 4\). This function effectively removes the sign of a number, leading to either zero or a positive number.
Why is this important in the context of the function \(f(x) = \frac{|x|}{x}\)? The numerator \(|x|\) evaluates to the positive value of \x\ regardless of whether \x\ is positive or negative. However, the denominator \x\ keeps its original sign.
This operation creates a piecewise function that is defined based on the sign of \x\, dictating different behavior for positive, negative, and zero inputs.

• If \x > 0\, both \|x|\ and \x\ are positive, so \(f(x) = 1\).
• If \x < 0\, \|x|\ is positive while \x\ is negative, resulting in \(f(x) = -1\).

Function Evaluation

Function evaluation involves finding the output of a function for specific inputs. In our exercise, this means using particular values for \x\ and observing how the function behaves.
The given function is \(f(x) = \frac{|x|}{x}\), a piecewise function that changes based on the value of \x\. Let’s elaborate on the specific evaluations:

• For \(x = -2\), the function returns \f(-2) = -1\ as \-2\ is negative.
• For \(x = 0\), \f(0)\ is undefined since you can't divide by zero.
• For \(x = 5\), the function becomes \f(5) = 1\ because 5 is positive.

These evaluations showcase the practical use of calculating the function output using different \x\ values.

Division by Zero

Division by zero is a rule in mathematics that represents operations that are undefined, mainly because it doesn’t create a valid or useful result.
When you divide a number by zero, you're essentially trying to determine how many times zero can fit into a certain number, which is not mathematically possible. That's why, in functions like \(f(x) = \frac{|x|}{x}\), if \x = 0\, the function becomes undefined.
Division by zero is important to recognize because it indicates boundaries or "breaking points" in functions. It often disrupts continuity and defines whether a function is complete or requires special definition at certain points.

• In \f(x)\, any situation where \x = 0\ causes a division by zero and, thus, the function doesn’t yield a valid number.

Function Inputs

Function inputs are the values you substitute into a function to receive an output. Every function is defined by rules that stipulate how to handle different kinds of inputs.
For our problem, we consider several inputs such as negative numbers, zero, positive numbers, and even expressions like \(x^2\) and \(\frac{1}{x}\). All these must be evaluated according to their particular characteristics.
Special input cases like \(f(x^2)\) and \(f(\frac{1}{x})\) show that inputs themselves can be expressions or functions.

• \(f(x^2)\) evaluates to 1 for all \x\ not equal to zero, since \(x^2\) is always non-negative, except at zero where it becomes undefined.
• For \(f(\frac{1}{x})\), the evaluation relies on the sign of \(\frac{1}{x}\), and at \x = 0\, this input becomes undefined too.

Always keep in mind the nature of the inputs because they significantly influence the output of the function.