Question

Show that the function \(y(x)=\exp (-|x|)\) defined by $$y(x)= \begin{cases}\exp x & \text { for } x<0 \\ 1 & \text { for } x=0 \\\ \exp (-x) & \text { for } x>0\end{cases}$$ is not differentiable at \(x=0 .\) Consider the limiting process for both \(\Delta x>0\) and \(\Delta x<0\)

Short answer

The left-hand limit of the derivative is 1, and the right-hand limit of the derivative is -1. Since these limits are not equal, the function is not differentiable at \(x = 0\).

Step-by-step solution

Define the function

The function given is:\[y(x) = \begin{cases}\exp(x) & \text{for } x < 0 \ 1 & \text{for } x = 0 \ \exp(-x) & \text{for } x > 0\end{cases}\]

Calculate the right-hand limit of the derivative

The right-hand limit as \(x \to 0^+\) is given by:\[\lim_{\Delta x \to 0^+} \frac{y(\Delta x) - y(0)}{\Delta x}\]Substitute \(y(x) = \exp(-x)\) for \(x > 0\) and \(y(0) = 1\):\[\lim_{\Delta x \to 0^+} \frac{\exp(-\Delta x) - 1}{\Delta x}\]Using the series expansion for \( \exp(-\Delta x) \):\[\exp(-\Delta x) \approx 1 - \Delta x + \frac{(-\Delta x)^2}{2!} + \cdots \]Thus:\[\frac{(1 - \Delta x + \frac{(-\Delta x)^2}{2!} + \cdots) - 1}{\Delta x} = \frac{-\Delta x + \frac{(- \Delta x)^2}{2!} + \cdots}{\Delta x}\]Simplifying, we get:\[\frac{- \Delta x}{\Delta x} = -1\]So, the right-hand limit of the derivative as \(\Delta x \to 0^+\) is \(-1\).

Calculate the left-hand limit of the derivative

The left-hand limit as \(x \to 0^-\) is given by:\[\lim_{\Delta x \to 0^-} \frac{y(\Delta x) - y(0)}{\Delta x}\]Substitute \(y(x) = \exp(x)\) for \(x < 0\) and \(y(0) = 1\):\[\lim_{\Delta x \to 0^-} \frac{\exp(\Delta x) - 1}{\Delta x}\]Using the series expansion for \( \exp(\Delta x) \):\[\exp(\Delta x) \approx 1 + \Delta x + \frac{(\Delta x)^2}{2!} + \cdots \]Thus:\[\frac{(1 + \Delta x + \frac{(\Delta x)^2}{2!} + \cdots) - 1}{\Delta x} = \frac{\Delta x + \frac{(\Delta x)^2}{2!} + \cdots}{\Delta x}\]Simplifying, we get:\[\frac{\Delta x}{\Delta x} = 1\]So, the left-hand limit of the derivative as \(\Delta x \to 0^-\) is \(1\).

Determine differentiability

A function is differentiable at a point if the left-hand and right-hand limits of the derivative are equal at that point. Here, we have:\[\lim_{x \to 0^+} \frac{y(x) - y(0)}{x} = -1\]and\[\lim_{x \to 0^-} \frac{y(x) - y(0)}{x} = 1\]Since \(-1 eq 1\), the limits do not match. Therefore, the function \(y(x) = \exp(-|x|)\) is not differentiable at \(x = 0\).

Key concepts

Piecewise Function

A piecewise function is defined by multiple sub-functions, each of which applies to a particular interval of the main function's domain. In this exercise, the function is given by:
\[y(x) = \begin{cases}\exp(x) & \text{for } x < 0 \ 1 & \text{for } x = 0 \ \exp(-x) & \text{for } x > 0\end{cases}\]
Here, there are three pieces:

• \( y(x) = \exp(x) \) for \( x < 0 \)
• \( y(x) = 1 \) for \( x = 0 \)
• \( y(x) = \exp(-x) \) for \( x > 0 \)

Each section of the function has a different mathematical expression based on the value of \( x \). This division makes it easier to evaluate the function at various points but requires careful consideration when discussing properties like continuity and differentiability at the boundaries, such as \( x = 0 \).

Derivative Limits

The constraint for a function to be differentiable at a point is that the left-hand limit and the right-hand limit of the derivative must be equal. For this function, we analyze the derivative limits as follows. To calculate the right-hand limit of the derivative as \( x \to 0^+ \), we use:
\[\lim_{\Delta x \to 0^+} \frac{y(\Delta x) - y(0)}{\Delta x}\]
Using the piece \( y(x) = \exp(-x) \) for \( x > 0 \), the expression becomes:
\[\lim_{\Delta x \to 0^+} \frac{\exp(-\Delta x) - 1}{\Delta x}\]
Using the series expansion for \( \exp(-\Delta x) \):
\[\exp(-\Delta x) \approx 1 - \Delta x + \frac{(-\Delta x)^2}{2!} + \cdots\]
So, the simplified result is:
\[\frac{-\Delta x}{\Delta x} = -1\]
For the left-hand limit as \( x \to 0^- \), we have:
\[\lim_{\Delta x \to 0^-} \frac{y(\Delta x) - y(0)}{\Delta x}\]
Using the piece \( y(x) = \exp(x) \) for \( x < 0 \), the expression becomes:
\[\lim_{\Delta x \to 0^-} \frac{\exp(\Delta x) - 1}{\Delta x}\]
With series expansion for \( \exp(\Delta x) \):
\[\exp(\Delta x) \approx 1 + \Delta x + \frac{(\Delta x)^2}{2!} + \cdots\]
This simplifies to:
\[\frac{\Delta x}{\Delta x} = 1\]
Here, the left-hand and right-hand limits are different (1 and -1 respectively), which implies that the function is not differentiable at \( x = 0 \).

Series Expansion

Understanding the series expansion of exponential functions is crucial for solving problems involving limits, especially in pieces of piecewise functions. The series expansion of an exponential function \( \exp(x) \) around \( x = 0 \) is given by:
\[\exp(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\]
For \( \exp(-x) \), the series becomes:
\[\exp(-x) = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \cdots\]
Using these expansions helps in approximating the exponential functions in the limits of derivatives. In the step-by-step solution:\

• For \( \exp(-\Delta x) \), we have used: \( 1 - \Delta x + \frac{(-\Delta x)^2}{2!} + \cdots \)


• For \( \exp(\Delta x) \), we have: \( 1 + \Delta x + \frac{(\Delta x)^2}{2!} + \cdots \)

These series approximations allow us to determine the derivative limits accurately and show why the function is not differentiable at \( x = 0 \).

Continuity

Continuity is a necessary but not a sufficient condition for differentiability. A function is said to be continuous at a certain point if the left-hand limit and right-hand limit equal the function's value at that point. For our given piecewise function:
\[y(x) = \begin{cases}\exp(x) & \text{for } x < 0 \ 1 & \text{for } x = 0 \ \exp(-x) & \text{for } x > 0\end{cases}\]
We know that:

• \( \lim_{x \to 0^-} \exp(x) = \exp(0) = 1 \)
• \( \lim_{x \to 0^+} \exp(-x) = \exp(0) = 1 \)

Since both the left-hand limit and right-hand limit as \( x \to 0 \) equal 1, and \( y(0) = 1 \), the function is continuous at \( x = 0 \). However, even though the function is continuous, the difference in derivative limits at \( x = 0 \) demonstrates that continuity alone does not ensure differentiability. Hence, while \( y(x) \) is continuous at \( x = 0 \), it is not differentiable there.