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Chapter 8: Problem 37

Let \(f(x)=\left\\{\begin{array}{rr}x^{2}+4 x, & -3 \leq x \leq 0 \\ -\sin x, & 0

Short Answer

Expert verified
Statments (a) and (c) are true; (b) and (d) are false.

Step by step solution

01

Understanding the Function Definition

The given piecewise function has three distinct expressions for different intervals of \(x\): 1) \(f(x) = x^2 + 4x\) for \(-3 \leq x \leq 0\), 2) \(f(x) = -\sin x\) for \(0 < x \leq \frac{\pi}{2}\), and 3) \(f(x) = -\cos x - 1\) for \(\frac{\pi}{2} < x \leq \pi\). Understanding how these functions behave in their respective domains is crucial for analyzing the given statements.
02

Verifying Global Minima at x=-2

To determine if \(x = -2\) is the global minimum, evaluate \(f(x) = x^2 + 4x\) at \(x = -2\): \((-2)^2 + 4(-2) = 4 - 8 = -4\). Check the endpoints and critical points of other pieces to compare: At \(x = -3\), \(f(-3) = 9 - 12 = -3\). For \(-\sin x\), the minimum value is -1 occurring at \(x = \frac{\pi}{2}\). For \(-\cos x - 1\), the minimum occurs at \(x = \pi\) with value \(-1 - 1 = -2\). Thus, the global minimum value is \(-4\) at \(x = -2\), making statement (a) true.
03

Verifying Global Maxima at x=蟺

Calculate \(f(x)\) for \(x = \pi\) using \(-\cos x - 1\): \(-\cos(\pi) - 1 = 1 - 1 = 0\). Compare across different pieces: For \(x^2 + 4x\), evaluate at endpoints and maximum \(x = -3, 0\): At \(x = -3\), \(f(-3) = -3\), and at \(x = 0\), \(f(0) = 0\). For \(-\sin x\), maximum value is 0 at \(x = 0\). Therefore, \(x = \pi\) is not the point of global maxima because both \(x = 0\) and \(x = \pi\) yield the same value. Statement (b) is false as it ties with \(x = 0\).
04

Checking Differentiability at x=蟺/2

Differentiate the relevant sections: \(f(x) = x^2 + 4x\) gives derivative \(2x + 4\), and \(-\sin x\) differentiates to \(-\cos x\). Evaluate derivatives at \(x = \frac{\pi}{2}\): The left-hand derivative \(= -\cos(\frac{\pi}{2}) = 0\), right-hand derivative from \(-\cos x - 1\) is \(\sin x\) which at \(\frac{\pi}{2}\) is 1. Since derivatives do not match, \(f(x)\) is non-differentiable at \(x = \frac{\pi}{2}\). Thus, statement (c) is true.
05

Checking Continuity at x=0

Evaluate limits: For \(-3 \leq x \leq 0\) piece, \(\lim_{x \to 0^{-}} f(x) = f(0) = 0\). For \(0 < x \leq \frac{\pi}{2}\), \(\lim_{x \to 0^{+}} f(x) = 0\) as \(-\sin(0) = 0\). Since \(\lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{+}} f(x) = f(0)\), \(f(x)\) is continuous at \(x = 0\). Statement (d) is false.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiability
Finding if a function is differentiable at a point is crucial because it tells us whether the function has a tangent line (a clear slope) at that point. It helps us understand how smooth a function is. For our piecewise function, we need to look at the point \(x = \frac{\pi}{2}\). To analyze differentiability, we must calculate the derivatives (a fancy word for slope) from both sides of a point.
To do this:
  • For \(-3 \leq x \lt 0\), the derivative is \(2x + 4\). Right before reaching \(x = \frac{\pi}{2}\), the derivative is obtained from \(-\cos x\), giving us 0 at \(x = \frac{\pi}{2}\).
  • For \(\frac{\pi}{2} \lt x \leq \pi\), the function \(-\cos x - 1\) has a derivative of \(\sin x\). At \(x = \frac{\pi}{2}\), this derivative equals 1.
Since these derivatives do not match, the function is not differentiable at \(x = \frac{\pi}{2}\). Differentiability fails here because the slopes of the left and right sides are different.
Continuity
Continuity at a point simplifies to three conditions - the limit from the left, the limit from the right, and the value of the function at the point must all be the same. Checking for continuity at \(x = 0\) allows us to see if there's a sudden jump or interruption in the graph.
Here鈥檚 how to break it down:
  • First, look at the segment \(-3 \leq x \leq 0\): Calculate \(f(x)\) at \(0\) as \(0^2 + 4 \times 0 = 0\). So, \(f(0) = 0\).
  • Next, for \(0 \lt x \leq \frac{\pi}{2}\): \(\lim_{x \to 0^{+}} f(x) = -\sin(0) = 0\).
Since the limit as you approach from the left is \(0\), from the right is \(0\), and the value at the point is also \(0\), this means the function is continuous at \(x = 0\). Thus, there isn't any break in the graph at this point.
Global extrema
Global extrema refer to the highest or lowest points over the entire domain of a function. Global maxima and minima are essential for understanding the boundaries of a function's behavior.
When evaluating global extrema, consider each segment of the piecewise function. Here's how:
  • For \(-3 \leq x \leq 0\), calculate extrema using critical and end points: The minimum value is \(-4\) at \(x = -2\).
  • For \(0 \lt x \leq \frac{\pi}{2}\), the function values range between \(0\) and \(-1\).
  • Finally, for \(\frac{\pi}{2} \lt x \leq \pi\), \(f(x)\) achieves a minimum at \(x = \frac{\pi}\) with \(-2\), and maximum at \(\pi\) with \(0\).
The global minimum is indeed \(-4\) at \(x = -2\), meaning that this is the lowest point across all the pieces. The global maximum happens at both \(x=0\) and \(x=\pi\) with value \(0\), showing these are the tallest points the function reaches.

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