Problem 20 - Thomas Calculus

Question

In Exercises \(15-20,\) sketch the graph of each function and determine whether the function has any absolute extreme values on its domain. Explain how your answer is consistent with Theorem \(1 .\) $$ f(x)=\left\\{\begin{array}{ll}{x+1,} & {-1 \leq x<0} \\ {\cos x,} & {0 < x \leq \frac{\pi}{2}}\end{array}\right. $$

Step-by-Step Solution

Answer

No absolute extremes over the entire domain due to a discontinuity at \(x = 0\).

Step 1: Identify the Function Components

The given function is a piecewise function. It is defined as two separate pieces: 1. For \(-1 \leq x < 0\), the function is \(f(x) = x + 1\).2. For \(0 < x \leq \frac{\pi}{2}\), the function is \(f(x) = \cos x\).Both sections are defined over their respective intervals and will be graphed separately.

Step 2: Graph Each Piece of the Function

First, graph \(f(x) = x + 1\) on the interval \([-1, 0)\). This is a straight line starting at \((x, y) = (-1, 0)\) and approaches but does not include \((0, 1)\).Next, graph \(f(x) = \cos x\) on the interval \((0, \frac{\pi}{2}]\). This curve starts from slightly above \((0, 1)\) and ends at \((\frac{\pi}{2}, 0)\).

Step 3: Analyze Continuity and Extremes

Check the continuity at \(x = 0\). There is a break; \(f(x)\) is not defined at \(x = 0\) (not continuous across the boundary). Thus, any extreme values will be within each piece rather than at the endpoint.For \(-1 \leq x < 0\), the absolute minimum is \(f(-1) = 0\) and the absolute maximum approaches \((0, 1)\) but isn't reached.For \(0 < x \leq \frac{\pi}{2}\), the function reaches an absolute maximum at \(f(0) \to 1\) and the minimum at \(f(\frac{\pi}{2}) = 0\).

Step 4: Consistency with Theorem 1

Theorem 1 in calculus generally states that an absolute maximum or minimum exists if the function is continuous on a closed interval. The function \(f(x)\) is not continuous over the entire interval \([-1, \frac{\pi}{2}]\) due to the discontinuity at \(x = 0\). This allows us to say that absolute extremes are considered separately per subinterval, hence finding potential extremes within separate subintervals as seen in Step 3.

Key Concepts

Extreme Values in Piecewise Functions

Piecewise functions can be tricky when it comes to identifying their extreme values--that is, the highest and lowest points on the graph. Given a piecewise function, it's essential to consider each piece separately. For each segment of the function:

• The absolute maximum is the largest y-value achieved within its range.
• The absolute minimum is the smallest y-value.

With our example function:- **For \(-1 \leq x < 0\):** The part of the function given is a linear portion, \(f(x) = x + 1\), which starts at \(f(-1) = 0\). As \(x\) gets closer to \(-1\), the function increases linearly toward \(f(0) \, \text{approaching}\, 1\), but this max isn't actually reached due to the open interval notation.
- **For \(0 < x \leq \pi / 2\):** In this range, \(f(x) = \cos x\) oscillates from nearly \(f(0) \to 1\) down to \(f(\frac{\pi}{2}) = 0\), showcasing an absolute max at nearly 1 and an actual minimum at 0.
Always observe that due to discontinuities, extremes might never be reached between parts of a function, thus calculated per segment.

Graphing Piecewise Functions

Graphing piecewise functions involves sketching each segment over its defined interval and considering endpoints carefully.1. **Identify Segments and Intervals:** - Divide the graphing task into pieces. Each section corresponds to a different equation and interval. - Highlight shifts in equations and note if intervals are closed or open, as these dictate whether endpoints are included.
2. **Graph Each Section:** - For the function \(f(x) = x + 1\) over \([-1, 0)\), plot a line from \((-1, 0)\) moving upwards to just shy of \(0, 1)\), essentially ending the graphing with an open circle at \((0, 1)\). - For \(f(x) = \cos x\) over \(0, \pi/2]\), start the curve just above \(0, 1\) heading toward \((\pi/2, 0)\). This segment indicates a steady decline in \(f\) values.
3. **Check Continuity:** - Examine intersection points, noting that a break occurs at \(x = 0\), where the graph jumps from a linear positive slope to beginning a cosine curve. Drawing piecewise functions precisely helps highlight any major shifts, like discontinuities or sudden changes in direction.

Continuity in Piecewise Functions

Continuity is a vital concept in understanding how a piecewise function behaves over its domain. To check this:

• Continuity implies no abrupt jumps; the function's graph can be drawn without lifting your pencil.
• Evaluate limits at points where pieces of the function meet.

Our function shows a notable break in continuity at \(x = 0\). Here's why:- **Discontinuity at \(x = 0\):** The first segment, \(f(x) = x + 1\), stops just before \(x = 0\), ending at an open point, whereas the second segment \(f(x) = \cos x\), starts just past it.- **Effect of Discontinuity:** Due to the jump, the function does not reach defined maximum or minimum values across \([-1, \frac{\pi}{2}]\) as a whole interval but instead only within subintervals.
Recognizing discontinuities aids in identifying where the potential for absolute extremes lies, and how the function needs to be analyzed segment by segment.