Problem 88 - University Calculus: Early Transcendentals

Question

$$\text { Find } f^{\prime}(0) \text { for } f(x)=\left\\{\begin{array}{ll}e^{-1 / x^{2}}, & x \neq 0 \\\0, & x=0\end{array}\right.$$.

Step-by-Step Solution

Answer

The derivative \( f'(0) = 0 \).

Step 1: Understanding the Function

The function given is a piecewise function. For any non-zero value of \( x \), the function is \( f(x) = e^{-1/x^2} \). At \( x = 0 \), the function is defined as \( f(0) = 0 \). Our task is to find the derivative of this function at \( x = 0 \).

Step 2: Use of Derivative Definition

To find \( f'(0) \), we use the definition of the derivative: \[ f'(x) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} \]. For \( x = 0 \), this becomes \[ f'(0) = \lim_{h \to 0} \frac{f(h) - 0}{h} = \lim_{h \to 0} \frac{e^{-1/h^2}}{h} \].

Step 3: Analyze the Limit Behavior

Consider the behavior of \( e^{-1/h^2} \) as \( h \to 0 \). Since \( -1/h^2 \) approaches negative infinity, \( e^{-1/h^2} \) approaches zero. Thus, the expression \( \frac{e^{-1/h^2}}{h} \) approaches zero as \( h \to 0 \).

Step 4: Conclusion on the Derivative

Since the limit of \( \frac{e^{-1/h^2}}{h} \) as \( h \to 0 \) is zero, we conclude that the derivative \( f'(0) = 0 \). Therefore, \( f'(0) \) is 0.

Key Concepts

Piecewise Functions

In mathematics, a piecewise function is a function that is defined by different expressions based on the value of the input. These functions are particularly useful when a situation changes behavior at certain points, requiring different expressions. For the given problem, the function

• is defined as \( f(x) = e^{-1/x^2} \) when \( x eq 0 \)
• and as \( f(x) = 0 \) when \( x = 0 \).

You can think of piecewise functions as being "piecemeal," dealing with different "pieces" of the function separately. They are managed by evaluating the different formulas depending on the input.
When dealing with derivatives of piecewise functions, it becomes essential to ensure that all pieces fit together smoothly.

Limit

Limits help us understand the behavior of a function as the input approaches a particular value. In our exercise, we need the limit to find the derivative at \( x = 0 \). The derivative formula requires a limit to capture how the function behaves around the point of interest.
In this problem, we calculated\[ f'(0) = \lim_{h \to 0} \frac{e^{-1/h^2}}{h} \]Here, as \( h \) gets extremely close to zero, the expression \( e^{-1/h^2} \) approaches zero because the exponent \( -1/h^2 \) becomes a large negative number making the whole expression very small.
Understanding limits is crucial when we deal with derivatives because they allow us to rigorously analyze the behavior of functions at specific points, giving rise to precise calculations of rates of change.

Exponential Function

Exponential functions take the form \( a^x \), but in our specific scenario, it's \( e^{-1/x^2} \), where \( e \) is the base. This powerful constant, approximately equal to 2.718, is crucial in diverse mathematical analyses due to its unique properties. In our problem, this exponential function changes very rapidly as \( x \) moves away from zero.
When \( h \) heads towards zero in \( e^{-1/h^2} \), the exponent \( -1/h^2 \) skyrockets to large negative values. This makes the value of \( e^{-1/h^2} \) approach zero because the exponential of a very large negative number approximates to zero.
So, with exponential functions, understanding their rate of growth or decay, especially with unique exponents, becomes vital in our analysis.

Continuity

Continuity is about whether a function has no breaks, jumps, or holes at a point or over an interval. For a piecewise function, checking continuity is crucial, particularly at points where the function's formula changes.
In our function, at \( x = 0 \), we observe the following:

• The limit of \( f(x) \) as \( x \to 0 \) from both sides is zero.
• Thus, \( f(x) \) is continuous at \( x = 0 \).

This continuity influences how we calculate derivatives because a continuous function implies a gentle transition between different parts. Hence, it ensures no abrupt changes in the slope, which is essential for understanding the behavior of derivatives in such contexts.