Chapter 2: Problem 9
Graph the function. 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 right-hand limit exist? $$f(x)=\left\\{\begin{array}{ll}\sqrt{1-x^{2}}, & 0 \leq x<1 \\\1, & 1 \leq x<2 \\\2, & x=2\end{array}\right.$$
Short Answer
Step by step solution
Analyze the Function
Graph the Function
Determine the Domain and Range
Identify Points of Existing Limits
Points with Only Left-Hand Limit
Points with Only Right-Hand Limit
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limits of Functions
For example, in our function, if we approach a point from the left side, we refer to the left-hand limit \(\lim_{x \to c^-} f(x)\), and from the right, the right-hand limit \(\lim_{x \to c^+} f(x)\).
In the given problem:
- From \(x = 0\) to \(x = 1\), the semicircle is continuous, meaning the limit exists for this interval as each value before 1 smoothly leads into the next.
- At \(x = 1\), though, we have a transition from the semicircle to a constant line, causing a disruption in continuity. Here, \(\lim_{x \to 1^-} f(x) = 0\) from the semicircle and \(\lim_{x \to 1^+} f(x) = 1\) from the constant function, hence the overall limit does not exist.
- At \(x = 2\), we encounter another gap since \(\lim_{x \to 2^-} f(x) = 1\) but \(f(2) = 2\), creating a discontinuity.
Domain and Range
In our case, the domain is \[ [0, 2] \] because the function covers a semicircle, a line segment, and includes a specific point at \(x = 2\). The entire span from 0 to 2 is covered, though certain points like \(x = 1\) and \(x = 2\) have unique definitions.
- For \(0 \leq x < 1\), it lies on a semicircle with \(f(x) = \sqrt{1-x^2}\).
- For \(1 \leq x < 2\), \(f(x) = 1\) over the straight line segment.
- Directly at \(x = 2\), \(f(x) = 2\), creating a single isolated point on the graph.
Graphing Functions
Such translation involves sketching:
- A semicircle, rising from \(x = 0, y = 1\) and dropping to \(x = 1, y = 0\). Don't connect \(x = 1\) because it's a separate component.
- A horizontal line at \(y = 1\) from \(x = 1\) to just before \(x = 2\), and finally,
- A singular point at \(x = 2, y = 2\), indicating a distinct value at this point.
Continuity of Functions
- The segment between \(x = 0\) and \(x < 1\) is continuous because the semicircle does not face any breaks.
- At \(x = 1\), there's a switch from the semicircle to a line, creating a jump and resulting in discontinuity. There's no seamless transition since \(\lim_{x \to 1^-} f(x) eq \lim_{x \to 1^+} f(x)\).
- Similarly, at \(x = 2\), the function jumps from the value on the line \(\lim_{x \to 2^-} = 1\) directly to \(f(2) = 2\), marking another discontinuity.