Problem 9 - Thomas Calculus

Question

Graph the functions in Exercises 9 and \(10 .\) Then answer these questions. a. What are the domain and range of \(f ?\) b. At what points \(c,\) if any, does \(\lim _{x \rightarrow c} f(x)\) exist? c. At what points does only the left-hand limit exist? d. At what points does only the left-hand limit exist? $$f(x)=\left\\{\begin{array}{ll}{\sqrt{1-x^{2}},} & {0 \leq x<1} \\ {1,} & {1 \leq x<2} \\ {2,} & {x=2}\end{array}\right.$$

Step-by-Step Solution

Answer

Domain: [0, 2]; Range: [0, 1] ∪ {2}. Limits exist at c ∈ (0, 1) and c=1. No exclusive left-hand limits.

Step 1: Graph the Function

First, we will graph the piecewise function. We divide the function into its respective intervals and parts:1. For \(0 \leq x < 1\), the function \(f(x) = \sqrt{1-x^{2}}\) represents the upper semicircle of the unit circle centered at the origin. This part of the function forms an arc from \(y=1\) to \(y=0\) as \(x\) goes from \(0\) to \(1\) (not including \(1\)).2. For \(1 \leq x < 2\), \(f(x) = 1\). This is a horizontal line including \(x=1\) (closed circle) but not \(x=2\) (open circle).3. At \(x=2\), \(f(x)=2\). This is just a point at \((2,2)\).

Step 2: Domain and Range of f

Now, let's find the domain and range of \(f\).- **Domain**: The domain of \(f\) is the set of all \(x\) for which \(f(x)\) is defined. Thus, it is \([0, 2]\) since \(f(x)\) is defined for all \(x\) in this interval.- **Range**: The range is the set of possible output values. From the graph, we see that \(f(x)\) values can include all values from \(0\) to \(1\) (due to the semicircle), and the specific value \(2\). Thus, the range is \([0, 1] \cup \{2\}\).

Step 3: Existence of Limits

For each point \(c\), check if the limit \(\lim_{x \to c} f(x)\) exists using the graph and its continuity:- From \(0 \leq x < 1\), the function \(\sqrt{1-x^{2}}\) is continuous, hence the limit exists for all \(c\) within \(0 < x < 1\).- At \(x=1\), the limit does exist because both the left-hand and right-hand limits reach \(1\).- At \(x=2\), the limit does not exist because the left-hand limit gives \(1\), whereas the point value is \(2\).Thus, the limits exist at all points of continuity \(0 < x < 1\) and at \(c=1\).

Step 4: Left-Hand Limit Existence

To find where only the left-hand limit exists, check for points where the function is connected from the left, but not to the right.- At \(x=1\), the left limit is \(1\) from the semicircle but connects smoothly to the value of 1.- The question asks for existence of only the left-hand limits, so no such special points are noted since both one-sided limits either exist equally or not at all separately apart from known discontinuities.Existing left-hand limits don't apply separately in usual continuity here.

Key Concepts

Limits and Continuity

A piecewise function can have sections where it is continuous and sections where it is not. Continuity means no breaks or jumps in the graph. In this exercise, let's track how the limits work in these contexts.

- For the interval \(0 \leq x < 1\), we have the upper semicircle graph represented by \(f(x) = \sqrt{1-x^{2}}\). It is smooth and continuous, meaning the limit as \(x\) approaches any point from within this segment exists. No jumps mean smooth sailing here!

- At \(x = 1\), something intriguing happens. The function moves from the semicircle to a horizontal line. However, both the left-hand and right-hand limits are equal here. That means \(\lim_{x \to 1} f(x) = 1\), showing that the function smoothly transitions despite having different expressions on either side.

- When \(x = 2\), there is a standout point \((2,2)\) which results in a break in continuity. Here, the left-hand limit coming from the interval \(1 \leq x < 2\) is \(1\), but it jumps directly up to \(2\), making the overall limit not exist at this specific point.

Domain and Range

The domain of a function tells us all the possible input values, and the range gives us the output results.

- For our function, the **domain** is \[0, 2\], because the function is defined from \(x = 0\) up to \(x = 2\), including both these endpoints.

- To understand the **range**, take a look at the output values from the function's parts:

• For \(0 \leq x < 1\), the function \(f(x) = \sqrt{1-x^{2}}\) gives us values from \(0\) to \(1\), forming a continuous arc.
• The region \(1 \leq x < 2\) keeps \(f(x) = 1\), contributing a single value of \(1\) to the range.
• Finally, the pinpoint \(x = 2\) adds the value \(2\). So, the function ups its count with another specific number, \(2\).

Hence, the complete range we see is \[0, 1\] \cup \{2\},\ meaning the curve stretches smoothly from 0 up to 1, and then includes just the exact value 2.

Graphing Functions

Graphing piecewise functions helps visualize their behavior across different sections.

- Start by plotting \(f(x) = \sqrt{1-x^{2}}\) from \(0 \leq x < 1\). Imagine this as a semicircle: it will arc from point \(y=1\) gradually down to \(y=0\). This is the "curvy" part of our graph!

- Next, the segment for \(1 \leq x < 2\) is simply \(f(x) = 1\). Draw this as a straight horizontal line extending to almost \(x = 2\), where it meets the axis.

- Finally, make a clear mark at \(x = 2\) for \((2,2)\). This is often represented as a distinct point sitting attractively on its own above the \(x = 2\) spot. It's essential to differentiate this separate point from the line and curve by using a filled circle or similar notation. This way, students can see where the function changes its way at the "joint" points, showing how the function transitions and where it might break in continuity.