Problem 25 - Contemporary Precalculus

Question

Show that the function \(f(x)=|x|+|x-2|\) is constant on the interval \([0,2] .\) [ Hint: Use the definition of absolute value (see Example \(7 \text { ) to compute } f(x) \text { when } 0 \leq x \leq 2 .]\)

Step-by-Step Solution

Answer

Question: Show that the function \(f(x) = |x| + |x - 2|\) is constant on the interval \([0, 2]\). Answer: For all cases in the given interval \([0, 2]\), we have computed the value of the function \(f(x) = |x| + |x - 2|\) and found that it is equal to \(2\). This shows that the function is constant on the interval, as required.

Step 1: Define the absolute value

The absolute value of a number is its distance from 0. Mathematically, we can define it as follows: $$ |x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases} $$

Step 2: Split the interval into cases

Since we need to determine the function's behavior on the interval \([0, 2]\), let's split the interval into cases and compute the value of the function in each case: 1. Case 1: \(0 \leq x < 1\) 2. Case 2: \(1 \leq x < 2\) 3. Case 3: \(x = 2\)

Step 3: Analyze Case 1 (\(0 \leq x < 1\))

In this case, \(x\) is between \(0\) and \(1\). Therefore, \(x \geq 0\) and \(x - 2 < 0\). Based on the definition of absolute value, we can find \(f(x)\) as follows: $$ f(x) = |x| + |x - 2| = x - (x - 2) = 2 $$ As we can see, for all values of \(x\) in this case, \(f(x) = 2\).

Step 4: Analyze Case 2 (\(1 \leq x < 2\))

In this case, \(x\) is between \(1\) and \(2\). Therefore, \(x \geq 0\) and \(0 \leq x - 2 < 1\). Based on the definition of absolute value, we can find \(f(x)\) as follows: $$ f(x) = |x| + |x - 2| = x + (x - 2) = 2x - 2 $$ Now, since \(1 \leq x < 2\), we know that \(2 < 2x < 4\). This implies that: $$ 0 < 2x - 2 < 2 $$ Therefore, for all values of \(x\) in this case, \(f(x) = 2\).

Step 5: Analyze Case 3 (\(x = 2\))

In this case, \(x\) is equal to \(2\). Then, \(x \geq 0\) and \(x - 2 = 0\). Using the definition of absolute value, we can find \(f(x)\) as follows: $$ f(x) = |x| + |x - 2| = 2 + 0 = 2 $$ Thus, for \(x = 2\), \(f(x) = 2\).

Step 6: Conclusion

For all cases in the given interval \([0, 2]\), we have computed the value of the function \(f(x) = |x| + |x - 2|\) and found that it is equal to \(2\). This shows that the function is constant on the interval, as required.

Key Concepts

Precalculus Foundations

Precalculus serves as the bridge between basic algebraic concepts and the more challenging realms of calculus. It encompasses various mathematical topics that provide the necessary groundwork for understanding complex functions, limits, and derivatives. One particularly important subset of precalculus is the study of piecewise functions and absolute value equations, which are fundamental to understanding how functions behave in different intervals.

A solid grasp of precalculus can be seen as assembling a toolkit that will be indispensable when we move on to calculus proper. Every tool – whether it’s factoring complex expressions, understanding the Cartesian coordinate system, or unraveling the intricacies of trigonometric functions – prepares the student to interpret and solve real-world problems using mathematical principles.

• Function behavior and analysis
• Trigonometric identities and functions
• Systems of equations and inequalities

Piecewise Functions

Piecewise functions are distinguished by their multiple expressions, each corresponding to different parts of the domain. They enable us to define functions that have distinct rules for different intervals. When plotting a piecewise function on a graph, it may appear as though the graph is 'broken up', which is an immediate visual cue of a piecewise nature of a function.

Understanding piecewise functions is crucial as they appear in a variety of mathematical and real-world situations, ranging from calculating taxes to describing the motion of an object under different conditions. When working with piecewise functions, it’s vital to:

• Determine the correct interval for the given input.
• Apply the function rule corresponding to that interval.
• Analyze the behavior of the function across its domain.

Example: Step Functions

Step functions are a classic example of piecewise functions. They remain constant within each subinterval, jumping to different values as the input crosses predefined thresholds.

Absolute Value Equations

The absolute value represents the distance of a number from zero on a number line, regardless of direction. Solving absolute value equations often involves splitting the problem into two separate cases – one where the variable is positive or zero, and the other where it is negative. This 'case analysis' approach is a powerful technique for tackling absolute value problems, as it allows us to dismantle the equations into simpler parts that are easier to solve.

When it comes to piecewise functions involving absolute values, this case analysis is essential. It allows us to create a clear picture of how the function behaves within different intervals, just as we have seen in the exercise where the function is evaluated in different intervals to determine its constancy. Key points to remember when dealing with absolute value equations include:

• The absolute value of a negative number is the number's positive counterpart.
• An equation with an absolute value may yield more than one solution.
• Graphical representations of absolute value equations typically produce 'V' shapes, reflecting their equidistance from the origin.