Question

\(f(x)=\max \\{2-x, 2+x, 4\\} x \in R\) is (a) \(f(x)=\mid \begin{array}{cc}2-x & x \geq 2 \\ 4 & -2

Short answer

The correct piecewise definition for the given function is: \(f(x) = \left\{ \begin{array}{cc} 2+x & x \leq -2 \\ 4 & -2 < x < 2 \\ 2-x & x \geq 2 \end{array} \right. \)

Step-by-step solution

Check the endpoints of the intervals

We have the intervals: \(x \leq -2, -2 < x < 2,\) and \(x \geq 2\). Let's check the endpoints \(x = -2\) and \(x = 2\). For \(x = -2\): \( 2-x = 2-(-2) = 4 \\ 2+x = 2+(-2) = 0 \\ \) For \(x = 2\): \( 2-x = 2-2 = 0 \\ 2+x = 2+2 = 4 \\ \)

Determine the correct function forms for each interval

Now, let's determine the highest function form for each interval. \(x \leq -2\): Considering the function values at the endpoint \(x = -2\), we saw that \(2-x = 4 = \max \{4, 0\}\). Therefore, the function form in this interval is \(2+x\). \(-2 < x < 2\): In this interval, the function forms \(2-x\) and \(2+x\) are both increasing from their respective endpoints, so they cannot be the highest simultaneously. As \(4\) is constant, we know that the maximum value in this interval is \(f(x) = 4\). \(x \geq 2\): Considering the function values at the endpoint \(x = 2\), we saw that \(2+x = 4 = \max \{0,4\}\). Therefore, the function form in this interval is \(2-x\).

Find the correct match among the given options

Now that we have found the correct function forms for each interval, we can look at the given options and determine which one matches our result. Our result: \( f(x) = \left\{ \begin{array}{cc} 2+x & x \leq -2 \\ 4 & -2 < x < 2 \\ 2-x & x \geq 2 \end{array} \right. \) Comparing it with the given options, we can see that option (d) is the correct one. So, the correct piecewise definition for the function is: \(f(x) = \left\{ \begin{array}{cc} 2-x & x \leq -2 \\ 4 & -2 < x < 2 \\ 2+x & x \geq 2 \end{array} \right. \)

Key concepts

Function Intervals

When dealing with piecewise functions, understanding function intervals is essential. Intervals define the range of values where specific parts of the function are applied. This means each interval corresponds to a different formula based on conditions tied to the variable, such as \(x\).For example, consider the exercise where \(f(x) = \max \{2-x, 2+x, 4\}\). It's broken into sections based on these intervals: \(x \leq -2\), \(-2 < x < 2\), and \(x \geq 2\).

• The interval \(x \leq -2\) signifies that we apply a certain formula when \(x\) is less than or equal to -2.
• In \(-2 < x < 2\), another formula is used for values of \(x\) strictly between -2 and 2.
• Lastly, the interval \(x \geq 2\) applies when \(x\) values are greater than or equal to 2.

By understanding these intervals, students can determine which part of the piecewise function to apply depending on the value of \(x\). This segmentation ensures the function accurately represents different patterns of behavior across its domain.

Maximum Value

The concept of the maximum value in this context refers to identifying the largest value among a set of expressions. For a function defined by \(f(x) = \max \{2-x, 2+x, 4\}\), the task is to find the largest numerical outcome from these expressions, given specific intervals.

• In the \(x \leq -2\) interval, calculations at the endpoint reveal that \((2-x)\) provides the maximum value of 4.
• For the interval \(-2 < x < 2\), while both \(2-x\) and \(2+x\) are increasing or decreasing functions, \(4\) remains constant and thus is the maximum value throughout.
• Finally, in the \(x \geq 2\) interval, \(2+x\) again reaches the value 4, making it the dominant expression that achieves the maximum.

Determining the maximum value helps complete the definition of the piecewise function, ensuring each interval is represented by the expression that maximizes the value.

Endpoints Evaluation

Endpoints evaluation involves examining the values of a function at the boundaries of its intervals. This is crucial for confirming which part of the function describes the maximum behavior.To analyze the endpoints in this specific problem:- At \(x = -2\): Verify the outcomes of each expression. \(2-x\) equals 4, while \(2+x\) equals 0. Thus, \(2-x\) is preferable as it provides a greater value.- At \(x = 2\): Re-check the expression outcomes. Here, \(2-x\) results in 0, and \(2+x\) results in 4, so \(2+x\) provides the greater output.By evaluating endpoints, students ensure accurate predictions of the function’s values at the critical junctions of each interval. The endpoint assessment confirms how the function behaves as boundaries change, allowing accurate conclusions for each interval in the piecewise definition.