Question

\(\mathrm{f}(\mathrm{x})=|\mathrm{x}+2|+|\mathrm{x}-1|\) is (a) increasing in \((-1, \infty)\) (b) increasing in \([-1, \infty]\) (c) decreasing in \((-\infty,-2)\) (d) decreasing in \([-\infty,-2]\)

Short answer

The correct answer is (c) decreasing in \((-\infty, -2)\), as the function is decreasing for all values of \(x < -2\).

Step-by-step solution

Identify the Piecewise Function

Given that \(f(x) = |x+2| + |x-1|\), let's write the function as a piecewise function considering the expressions inside the absolute value. Let's define: \(g(x) = x + 2\) and \(h(x) = x - 1\) Now, we can write the piecewise function: \[ f(x) = \begin{cases} -g(x)-h(x) & x<-2 \\ -g(x)+h(x) & -2 \le x < 1 \\ g(x)+h(x) & x \ge 1 \end{cases} \]

Find the Derivatives

Now, we will find the derivatives for each case. 1. For the first case, where \(x < -2\), the derivative is: \(f'(x) = -g'(x) - h'(x)\) 2. For the second case, where \(-2 \le x < 1\), the derivative is: \(f'(x) = -g'(x) + h'(x)\) 3. For the third case, where \(x \ge 1\), the derivative is: \(f'(x) = g'(x) + h'(x)\) Now, we can find the derivatives for \(g'(x)\) and \(h'(x)\): \(g'(x) = 1\) and \(h'(x) = 1\)

Analyze the Derivatives

Now, let's analyze the signs of the derivatives for each case, which will let us know if the function is increasing or decreasing. 1. For the first case, where \(x < -2\), \(f'(x) = -1 - 1 = -2\). Since \(f'(x) < 0\), the function is decreasing. 2. For the second case, where \(-2 \le x < 1\), \(f'(x) = -1 + 1 = 0\). Since \(f'(x) = 0\), the function is neither increasing nor decreasing. 3. For the third case, where \(x \ge 1\), \(f'(x) = 1 + 1 = 2\). Since \(f'(x) > 0\), the function is increasing. Based on this analysis, we can conclude the correct answer from the given options. a) increasing in \((-1, \infty)\): This is not correct, because the function is only increasing for \(x \ge 1\). b) increasing in \([-1, \infty]\): This is not correct, because the function is only increasing for \(x \ge 1\). c) decreasing in \((-\infty, -2)\): This is the correct answer, as the function is decreasing for all values of \(x < -2\). d) decreasing in \([- \infty, -2]\): This is not correct, because the function is only decreasing for \(x < -2\).

Key concepts

Absolute Value Functions

Absolute value functions are unique and fascinating parts of mathematics. They have a distinctive kink where the function "bends," making them very different from linear equations. The absolute value of a number represents its non-negative magnitude regardless of its direction on the number line.

For example, with the absolute value function \( f(x) = |x+2| + |x-1| \), we need to consider the values inside the absolute value brackets separately. This is crucial because:

• When \(x + 2\) or \(x - 1\) is negative, the expression changes its sign.
• This creates a "piecewise" nature in the function, meaning it behaves differently across different sections of its domain.

This piecewise nature requires us to break the problem down into segments, thus helping to reveal the true behavior of the function at specific intervals.

Derivative Analysis

Understanding derivative analysis helps us study how functions change. Taking the derivative involves finding the rate at which something changes. For absolute value functions like \( f(x) = |x+2| + |x-1| \), this is especially interesting.

We first express the function in a piecewise manner, and then look at each segment individually. By finding the derivative for each piece, we can determine how the original function behaves:

• When we look at \( f'(x) \), it tells us the function's slope: whether it rises or falls or remains flat over a specific interval.
• In this exercise, we calculate \( f'(x) \) separately for intervals \( x < -2 \), \( -2 \le x < 1 \), and \( x \ge 1 \).

Hence, the derivative provides critical insight into the function's local and global behavior, helping us predict increases or decreases.

Increasing and Decreasing Functions

Detecting if a function is increasing or decreasing is pivotal in mathematical analysis. This knowledge uncovers trends and behaviors.

For example, for the piecewise function \( f(x) = |x+2| + |x-1| \):

• If \( f'(x) > 0 \), then the function is increasing at that interval.
• If \( f'(x) < 0 \), the function decreases.
• Where \( f'(x) = 0 \), it's flat, meaning no "net" change occurs.

In our original problem:
- The function decreases where \( x < -2 \). This is depicted by \( f'(x) = -2 \).
- It remains steady between \( -2 \le x < 1 \), as \( f'(x) = 0 \).
- It increases when \( x \ge 1 \) since \( f'(x) = 2 \).
Like detective work, identifying these trends unveils a function's entire story.

Mathematical Problem Solving

Mathematical problem solving is an essential skill that involves structured exploration and analysis to reach conclusions. Approach exercises like these systematically for the best results. When dealing with absolute value functions:

1. **Break Down**: Divide complex functions into manageable pieces. This allows for focused analysis on each section. By defining segments like \( x < -2 \), \( -2 \le x < 1 \), and \( x \ge 1 \), we isolate key characteristics.

2. **Explore Derivatives**: Calculate derivatives for each section to pinpoint how the function changes. This step reveals important clues about increasing or decreasing tendencies.

3. **Analyze Trends**: Look for signs from derivatives to understand global behavior. This helps in verifying whether the function rises, falls, or stays constant across intervals.

By adopting a detailed and thorough approach, students can tackle challenging problems efficiently and with confidence, learning valuable skills along the way.