Chapter 2: Problem 19
Explain why the function is discontinuous at the given number \(a\) . Sketch the graph of the function. \(f(x)=\left\\{\begin{array}{ll}{\cos x} & {\text { if } x<0} \\ {0} & {\text { if } x=0} \\ {1-x^{2}} & {\text { if } x>0}\end{array}\right. \quad a=0\)
Short Answer
Expert verified
The function is discontinuous at \(x = 0\) because \(f(0) = 0\), but both left and right-hand limits are 1.
Step by step solution
01
Define Discontinuity
A function is discontinuous at a point if it lacks a value at that point or if there is a break, jump, or hole in the graph. In mathematical terms, for a point \(a\), a function \(f(x)\) is discontinuous at \(a\) if \(\lim_{x \to a^-} f(x) eq \lim_{x \to a^+} f(x)\), or if these limits exist but are not equal to \(f(a)\).
02
Calculate Left-Hand Limit
Evaluate the limit of \(f(x)\) as \(x\) approaches 0 from the left. For \(x < 0\), \(f(x) = \cos x\). Thus, \(\lim_{x \to 0^-} f(x) = \cos 0 = 1\).
03
Calculate Right-Hand Limit
Evaluate the limit of \(f(x)\) as \(x\) approaches 0 from the right. For \(x > 0\), \(f(x) = 1 - x^2\). Thus, \(\lim_{x \to 0^+} f(x) = 1 - 0^2 = 1\).
04
Evaluate the Function at \(x = 0\)
Since \(f(0) = 0\), because the function explicitly assigns this value when \(x = 0\).
05
Compare Limits and Function Value
Now, compare the left-hand limit, right-hand limit, and the function value at \(x = 0\). We have \(\lim_{x \to 0^-} f(x) = 1\), \(\lim_{x \to 0^+} f(x) = 1\), but \(f(0) = 0\). Since the limits do not match the function value, the function is discontinuous at \(x = 0\).
06
Sketch the Graph
Draw three pieces for the graph: 1. The graph of \(f(x) = \cos x\) for \(x < 0\), which oscillates between -1 and 1, getting closer to 1 as it approaches 0. 2. A point at \(x=0, y=0\) showing \(f(0)\). 3. The graph of \(f(x) = 1 - x^2\) for \(x > 0\), starting at \(y=1\) and creating a downward-opening parabola.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Piecewise Function
A piecewise function is a type of function defined by different expressions over different intervals of the input, typically separated by a dividing point. In this exercise, the function is defined separately for when the input is less than, equal to, or greater than zero. This is a classic case of a piecewise function where:
- For values of x less than 0, it uses the function \( \cos x \).
- For x exactly at 0, it is specified as 0.
- For x greater than 0, it follows the rule \( 1-x^2 \).
Limit Definition of Continuity
To understand if a function is continuous at a certain point, you must consider the function's limits. For continuity at a point, say \( a \), three conditions should be satisfied:
The left-hand limit is \( \cos 0 = 1 \), and the right-hand limit for \( x > 0 \) is also \( 1 - 0^2 = 1 \). Yet, the value of the function at \( x = 0 \) is given as 0. Since these limits do not match \( f(0) \), the function is discontinuous at \( x = 0 \). This demonstrates the necessity of all three conditions being satisfied for continuity.
- The left-hand limit \( \lim_{x \to a^-} f(x) \) exists.
- The right-hand limit \( \lim_{x \to a^+} f(x) \) exists.
- These limits must equal the exact output of the function at \( a \), thus \( \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a) \).
The left-hand limit is \( \cos 0 = 1 \), and the right-hand limit for \( x > 0 \) is also \( 1 - 0^2 = 1 \). Yet, the value of the function at \( x = 0 \) is given as 0. Since these limits do not match \( f(0) \), the function is discontinuous at \( x = 0 \). This demonstrates the necessity of all three conditions being satisfied for continuity.
Graph Sketching
Sketching the graph of a piecewise function involves plotting each segment according to its defined rule and considering any breaks in continuity. Start by sketching each segment of the function:
- For \( x < 0 \), plot the curve \( \cos x \), which approaches 1 as \( x \) nears 0 from the left.
- For \( x = 0 \), note the isolated point at \( (0, 0) \), representing \( f(0) = 0 \).
- For \( x > 0 \), draw the downward-opening parabola \( 1 - x^2 \), which starts at \( y = 1 \) as \( x \) begins to increase from 0.
Left-Hand and Right-Hand Limits
Understanding left-hand and right-hand limits is key to analyzing discontinuities. When approaching a point from the left, the left-hand limit describes the function's behavior, while approaching from the right, the right-hand limit comes into play.
For our function, as \( x \to 0^- \), or from the left, the limit is \( \cos 0 = 1 \), while from the right, as \( x \to 0^+ \), we compute \( 1 - 0^2 = 1 \).
Both limits equal 1, illustrating that the function behaves predictably from both sides. Yet, due to the discrete definition at \( x = 0 \), where \( f(0) = 0 \), a gap appears between \( f(0) \) and the limit values. This clear mismatch highlights why identifying these are vital to diagnosing discontinuities in a function.
For our function, as \( x \to 0^- \), or from the left, the limit is \( \cos 0 = 1 \), while from the right, as \( x \to 0^+ \), we compute \( 1 - 0^2 = 1 \).
Both limits equal 1, illustrating that the function behaves predictably from both sides. Yet, due to the discrete definition at \( x = 0 \), where \( f(0) = 0 \), a gap appears between \( f(0) \) and the limit values. This clear mismatch highlights why identifying these are vital to diagnosing discontinuities in a function.