Problem 37
Question
Sketch the graph of a function that has domain \([0,2]\) and is continuous on \([0,2)\) but not on \([0,2]\).
Step-by-Step Solution
Verified Answer
Sketch \(f(x) = x\) for \(x \in [0, 2)\) with \(f(2) = 1\).
1Step 1: Understand the Domain and Continuity Requirements
The function must be defined for all values in the interval \([0, 2]\) and remain continuous up to but not including the endpoint \(x = 2\). At \(x = 2\), the function should have a point of discontinuity.
2Step 2: Sketch a Continuous Segment on [0, 2)
Begin by sketching a graph on the interval \([0, 2)\) that is continuous. A simple example could be the function \(f(x) = x\), which is a linear function. Draw a line from \(x = 0\) to just before \(x = 2\) without including \(x = 2\).
3Step 3: Introduce Discontinuity at x = 2
To introduce the discontinuity at \(x = 2\), you could define the function just outside the continuous behavior. For instance, let \(f(2) = 1\), a different value than would be expected from the continuity in \([0, 2)\). Represent this with a point on the graph at the coordinate \((2, 1)\) that does not connect with the line from \([0, 2)\).
4Step 4: Verify Continuity and Discontinuity
Check that as \(x\) approaches \(2\) from the left, the function appears continuous (e.g., it approaches the expected value if \(f(x) = x\), approaching \(f(2) = 2\)), but at \(x = 2\) the function value changes to \(f(2) = 1\). This confirms discontinuity at \(x = 2\).
Key Concepts
GraphingDomainContinuityDiscontinuity
Graphing
Graphing helps us visually interpret and analyze the behavior of functions. When tackling an exercise like this, sketching provides a clear understanding of how the function behaves across its domain. For functions such as this one, with a specific continuity and discontinuity characteristic, start by drafting the section that is supposed to be continuous first.
- Identify what portions of the domain the function should remain continuous. Our given exercise specifies this for the interval \([0, 2)\).
- Draw the appropriate segment for the function, which could be as simple as a straight line, or a curve, depending on the function's nature.
- In our case, a line representing\(f(x) = x\) from \(x = 0\) to just before \(x = 2\) effectively captures the continuous part.
Domain
The domain is fundamental in defining the "universe of discourse" of a function, meaning it describes all possible values of \(x\) for which the function is defined. In our exercise, the function has a well-defined domain of \([0, 2]\). This includes every real number from 0 up to and including 2.
- Consider \([0, 2)\) as the portion where the function is continuous, meaning that every value in this interval naturally leads to its respective function output.
- The domain also includes the endpoint \(x = 2\), but as specified, there's a discontinuity right at this point.
Continuity
Continuity in functions describes how seamlessly one can move from one point to another within a function's domain without interruption. This is crucial in many mathematical applications, as continuous functions ensure predictability and smooth transitions.
- For a function to be continuous on an interval like \([0, 2)\), it must smoothly connect every value in that interval without any sudden jumps or breaks.
- In our example, using \(f(x) = x\) illustrates this as we start with \(x = 0\) and progresses linearly to just below \(x = 2\), without deviation from the expected path.
Discontinuity
Discontinuity represents a break, gap, or jump in the graph of a function, making it non-connected at a point or over an interval. Identifying discontinuities is crucial to understanding exceptions where the function's behavior differs from its typical pattern.
- In this exercise, the discontinuity occurs exactly at \(x = 2\), where the function \(f(x)\) should not "follow through" smoothly from the rest of the interval \([0, 2)\).
- Specifically, we assign \(f(2) = 1\), creating a deliberate divergence from the expected value (which would be \(2\) if it were continuous).
Other exercises in this chapter
Problem 37
In Problems 37-40, use natural logarithms to solve each of the exponential equations. Hint: To solve \(3^{x}=11\), take ln of both sides, obtaining \(x \ln 3=\l
View solution Problem 37
Find \(\lim _{x \rightarrow 1}\left(x^{2}-1\right) /|x-1|\) or state that it does not exist.
View solution Problem 37
Prove that \(\lim _{x \rightarrow c} f(x)=L \Leftrightarrow \lim _{x \rightarrow c}[f(x)-L]=0\).
View solution Problem 37
Find the limits. \(\lim _{x \rightarrow 0^{+}} \frac{[x]}{x}\)
View solution