Chapter 3: Problem 36
Does the graph of $$g(x)=\left\\{\begin{array}{ll}x \sin (1 / x), & x \neq 0 \\\0, & x=0\end{array}\right.$$ have a tangent at the origin? Give reasons for your answer.
Short Answer
Expert verified
The graph does not have a tangent at the origin.
Step by step solution
01
Understand the Piecewise Function
The function \( g(x) \) is defined piecewise. For \( x eq 0 \), the expression is \( g(x) = x \sin(1/x) \), which involves the sine function all wrapped with a fraction in the argument, making its behavior complex as we approach zero. At \( x = 0 \), the function is specifically defined as \( g(0) = 0 \).
02
Check Continuity at the Origin
For a function to be differentiable at a point, it must first be continuous there. As \( x \to 0 \), evaluate \( \lim_{x\to 0} g(x) = \lim_{x\to 0} x \sin(1/x) \). Since \( \sin(1/x) \) is bounded between -1 and 1, \( |x \sin(1/x)| \leq |x| \). Thus, \( \lim_{x\to 0} x \sin(1/x) = 0 \), confirming that \( g(x) \) is continuous at \( x = 0 \).
03
Evaluate Differentiability Using the Definition
Differentiability at a point \( x = a \) requires that \( \lim_{h\to 0} \frac{g(a+h) - g(a)}{h} \) exists. Here, let \( a = 0 \), so we need \( \lim_{h \to 0} \frac{h \sin(1/h) - 0}{h} = \lim_{h \to 0} \sin(1/h) \). The sine function oscillates between -1 and 1 as \( h \to 0 \), so the limit does not exist because the values do not settle to a single number.
04
Conclusion on the Existence of Tangent
Since the limit in Step 3 does not exist, \( g(x) \) is not differentiable at \( x = 0 \). Therefore, the graph does not have a tangent line at the origin.
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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 that is defined by different expressions depending on the input value or region. In our given example, the function \( g(x) \) is defined differently for \( x eq 0 \) and \( x = 0 \). For \( x eq 0 \), the expression is \( g(x) = x \sin(1/x) \). This part of the function can be complex as it involves trigonometric properties bundled within fractions.
At \( x = 0 \), the function is explicitly defined as \( g(0) = 0 \). This means for any other value except 0, the function behaves according to the formula given, but at the origin, it is fixed at a particular value to maintain definition and manage behavior close to zero. Understanding piecewise functions is crucial because they combine multiple expressions for more comprehensive modeling of real-world problems.
At \( x = 0 \), the function is explicitly defined as \( g(0) = 0 \). This means for any other value except 0, the function behaves according to the formula given, but at the origin, it is fixed at a particular value to maintain definition and manage behavior close to zero. Understanding piecewise functions is crucial because they combine multiple expressions for more comprehensive modeling of real-world problems.
Continuity
Continuity is a fundamental property that checks whether a function smoothly travels through points without any breaks or jumps. For the function \( g(x) \), we need to check if it is continuous at the origin, \( x = 0 \). To determine this, we evaluate \( \lim_{x \to 0} g(x) \).
Since \( g(x) = x \sin(1/x) \) for \( x eq 0 \), and knowing \( \sin(1/x) \) is oscillating but bounded between -1 and 1, we apply the squeeze theorem:\[ |x \sin(1/x)| \leq |x| \]
As \( x \to 0 \), \( |x| \to 0 \), hence \( \lim_{x \to 0} x \sin(1/x) = 0 \).
This shows that \( g(x) \) approaches 0 as \( x \to 0 \), which matches \( g(0) = 0 \). Thus, \( g(x) \) is continuous at \( x = 0 \), an essential step before checking differentiability.
Since \( g(x) = x \sin(1/x) \) for \( x eq 0 \), and knowing \( \sin(1/x) \) is oscillating but bounded between -1 and 1, we apply the squeeze theorem:\[ |x \sin(1/x)| \leq |x| \]
As \( x \to 0 \), \( |x| \to 0 \), hence \( \lim_{x \to 0} x \sin(1/x) = 0 \).
This shows that \( g(x) \) approaches 0 as \( x \to 0 \), which matches \( g(0) = 0 \). Thus, \( g(x) \) is continuous at \( x = 0 \), an essential step before checking differentiability.
Tangent Line
The concept of a tangent line revolves around the existence of a straight line that just touches a curve at a specific point, without crossing it. This line indicates the immediate rate of change of the function at that particular point. For a tangent line to exist at a point on a function, not only must the function be continuous at that point, but it also needs to be differentiable there.
In our case, examining whether \( g(x) \) has a tangent line at the origin entails checking the behavior of derivatives at \( x = 0 \). Since the continuity condition is satisfied, the next step is to explore the limit that forms the derivative definition at zero.
If the function can’t follow a unique path direction at the origin due to abrupt changes or oscillations (as is the case here), it means a tangent line cannot be formed at \( x = 0 \). Therefore, differentiability analysis is crucial for assessing the tangent.
In our case, examining whether \( g(x) \) has a tangent line at the origin entails checking the behavior of derivatives at \( x = 0 \). Since the continuity condition is satisfied, the next step is to explore the limit that forms the derivative definition at zero.
If the function can’t follow a unique path direction at the origin due to abrupt changes or oscillations (as is the case here), it means a tangent line cannot be formed at \( x = 0 \). Therefore, differentiability analysis is crucial for assessing the tangent.
Limit Evaluation
Limit evaluation is a critical process in analyzing the behavior of functions as they approach a particular point, especially for checking continuity and differentiability. When determining whether the function \( g(x) \) is differentiable at \( x = 0 \), we calculate the derivative limit:
\[ \lim_{h \to 0} \frac{g(h) - g(0)}{h} \]
Given \( g(h) = h \sin(1/h) \), the expression becomes \( \lim_{h \to 0} \sin(1/h) \). The sine function inherently oscillates between -1 and 1 as \( h \to 0 \). Such behavior means the limit doesn’t stabilize to a single value; hence, it diverges.
This absence of a fixed limit at \( x = 0 \) results in the function lacking differentiability there. So, no tangent line exists at that point.
\[ \lim_{h \to 0} \frac{g(h) - g(0)}{h} \]
Given \( g(h) = h \sin(1/h) \), the expression becomes \( \lim_{h \to 0} \sin(1/h) \). The sine function inherently oscillates between -1 and 1 as \( h \to 0 \). Such behavior means the limit doesn’t stabilize to a single value; hence, it diverges.
This absence of a fixed limit at \( x = 0 \) results in the function lacking differentiability there. So, no tangent line exists at that point.