Problem 88 - Thomas Calculus in SI Units

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 \( f'(0) \) is 0.

Step 1: Understand the Function

The function \( f(x) \) is defined as a piecewise function. For \( x eq 0 \), \( f(x) = e^{-1/x^2} \). At \( x = 0 \), \( f(x) = 0 \). We need to find \( f'(0) \), the derivative at \( x = 0 \).

Step 2: Definition of Derivative

The derivative of a function \( f(x) \) at a point \( a \) is given by the limit \( f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \). Here, we need \( f'(0) \), so \( a = 0 \).

Step 3: Write the Limit Expression for Derivative

Since \( f(0) = 0 \), the derivative \( f'(0) \) is computed as: \[ f'(0) = \lim_{h \to 0} \frac{f(h) - 0}{h} = \lim_{h \to 0} \frac{e^{-1/h^2}}{h}. \]

Step 4: Evaluate the Limit

As \( h \to 0 \), \( e^{-1/h^2} \) approaches zero very rapidly, because the exponent \( -1/h^2 \) tends to \(-\infty\). Therefore, \[ \lim_{h \to 0} \frac{e^{-1/h^2}}{h} = \lim_{h \to 0} \frac{0}{h} = 0. \]

Step 5: Conclusion

The limit evaluates to zero, thus the derivative of \( f(x) \) at \( x = 0 \) is zero. Therefore, \( f'(0) = 0 \).

Key Concepts

Derivative

A derivative represents the rate of change of a function concerning one of its variables. In simpler terms, it measures how fast or slow something changes. You can think of it as calculating the slope of the tangent line to a point on the graph of the function.

When we talk about derivatives in calculus, we're looking at the derivative at a particular point. To find this, we use a specific formula:

• Derivative at a point: \( f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \).

This formula tells us to take the limit as \( h \) approaches zero of the difference between the values of the function at \( a+h \) and \( a \), divided by \( h \). In this exercise, to find the derivative of \( f(x) \) at \( x = 0 \), we applied this formula to the piecewise function defined.

Piecewise Functions

Piecewise functions are functions that have different expressions based on the input value or interval. They allow for a function to behave in one way over one region and in another way over a different region.

In our exercise, the function \( f(x) \) is specified as:

• \( f(x) = e^{-1/x^2} \) for \( x eq 0 \)
• \( f(x) = 0 \) for \( x = 0 \)

This shows that the function has one expression when \( x \) is not equal to zero, and a different one when \( x \) is exactly zero.

Piecewise functions require careful handling, particularly at the boundaries between their intervals, since they can behave dramatically differently in those areas.

Limits

Limits help us understand the behavior of functions as they approach a certain point. They are foundational in calculus, often used to define derivatives and integrals.

In the derivative's context, limits are used to find the derivative at a particular point. The limit describes what happens to the value of a function as we get closer and closer to a specific value from either direction.

In this problem, evaluating the limit was crucial to determining the derivative at \( x = 0 \). By finding:

• \( \lim_{h \to 0} \frac{e^{-1/h^2}}{h} \)

we determined how \( f(x) \) approaches behavior as \( x \) nears zero. As \( h \) heads to zero, \( e^{-1/h^2} \) rapidly goes to zero, helping us conclude that the limit - and hence the derivative - is zero.

Understanding limits assists in handling functions at points where they might not be explicitly defined or have unusual behavior.