Problem 36 - University Calculus: Early Transcendentals

Question

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.

Step-by-Step Solution

Answer

The graph of \( g(x) \) does not have a tangent at the origin, as the derivative does not exist there.

Step 1: Understand the Problem

We are given a piecewise function \( g(x) \) and need to determine if the graph of \( g(x) \) has a tangent at the origin, \( x = 0 \). A tangent at a point is defined if the derivative exists at that point.

Step 2: Check the Differentiability at \( x=0 \)

To have a tangent, the function must be differentiable at the point in question. This means the derivative from both sides of \( x = 0 \) should exist and be equal.

Step 3: Differentiate \( g(x) \) for \( x \neq 0 \)

The expression for \( g(x) \) when \( x eq 0 \) is \( x \sin(1/x) \). We need to differentiate this. Use the product rule on \( g(x) = u(x)v(x) \) where \( u(x) = x \) and \( v(x) = \sin(1/x) \).

Step 4: Apply Product Rule

Using the product rule: \( (uv)' = u'v + uv' \). Here, \( u'(x) = 1 \) and \( v'(x) = \cos(1/x)(-1/x^2) \) (chain rule). Substituting, we get:\[g'(x) = 1 \cdot \sin(1/x) + x \cdot \cos(1/x) \cdot (-1/x^2)\]This simplifies to:\[g'(x) = \sin(1/x) - \cos(1/x)/x\]

Step 5: Evaluate the Limit as \( x \to 0 \)

The derivative at \( x = 0 \) is based on the limit:\[ \lim_{{x \to 0}} \frac{g(x) - g(0)}{x - 0} = \lim_{{x \to 0}} \frac{x \sin(1/x)}{x} = \lim_{{x \to 0}} \sin(1/x) \]However, since \( \sin(1/x) \) oscillates between \(-1\) and \(1\) as \( x \to 0 \), this limit does not exist.

Step 6: Conclude Differentiability at \( x=0 \)

Since the limit defining the derivative does not exist, \( g'(0) \) does not exist. Therefore, the function \( g(x) \) is not differentiable at \( x = 0 \), and thus, it does not have a tangent at the origin.

Key Concepts

Piecewise function

A piecewise function is a function that is defined by different expressions based on the input values or intervals. In simpler terms, it "pieces" together various functions to create a new one. For example, the function defined in the exercise, \( g(x) \), shows us two different properties:

• For \( x eq 0 \), \( g(x) = x \sin(1/x) \).
• For \( x = 0 \), \( g(x) = 0 \).

This setup allows the behavior of the function to vary depending on which interval \( x \) falls into. At a glance, piecewise functions might seem complicated, but they follow the same rules as other functions. When evaluating such functions or determining their derivatives, one must consider each piece separately.

Limit does not exist

The concept of limits is essential when discussing differentiability and continuity at certain points, especially for piecewise functions. A limit exists if, as you approach a specific point from both sides, the values you approach are the same. However, if the function behaves unpredictably or oscillates without approaching a single value, then the limit doesn't exist.

In the provided exercise, to find if the function \( g(x) \) has a tangent at the origin, we check the limit of the derivative as \( x \) approaches 0:

• The expression \( \lim_{{x \to 0}} \sin(1/x) \) is considered.
• Since \( \sin(1/x) \) oscillates rapidly between -1 and 1 as \( x \) nears 0, it does not settle to a single value.

Therefore, we say that the limit does not exist in this case. Without a limit, the derivative cannot be defined at that point, meaning there is no tangent at the origin.

Product rule

The product rule is a method used to find the derivative of a product of two functions. In calculus, when you have a function that can be expressed as the product of two simpler functions, the product rule is vital for differentiation.

Given two functions, \( u(x) \) and \( v(x) \), the derivative of their product, \( g(x) = u(x)v(x) \), is computed as:

• Apply the product rule formula: \[(uv)' = u'v + uv'\]
• In the exercise, \( g(x) = x \sin(1/x) \) is split into \( u(x) = x \) and \( v(x) = \sin(1/x) \).
• Calculate their derivatives: \( u'(x) = 1 \) and, applying the chain rule, \( v'(x) = \cos(1/x) (-1/x^2) \).
• Substitute into the product rule to construct the derivative: \[g'(x) = 1 \cdot \sin(1/x) + x \cdot \cos(1/x) \cdot (-1/x^2)\]
• Simplify: \[g'(x) = \sin(1/x) - \cos(1/x)/x\]

Mastering the product rule helps in differentiating more complex functions efficiently.