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Chapter 4: Problem 18

Determine whether Rolle's Theorem applies to the following functions on the given interval. If so, find the point(s) guaranteed to exist by Rolle's Theorem. $$h(x)=e^{-x^{2}} ;[-a, a],\( where \)a>0$$

Short Answer

Expert verified
Answer: Yes, Rolle's Theorem applies to the function \(h(x) = e^{-x^2}\) on the interval \([-a, a]\). The point c guaranteed by the theorem is \(c = 0\).

Step by step solution

01

Check Continuity on [a, b]

The given function is $$h(x) = e^{-x^2}.$$Since the exponential function is continuous everywhere on its domain, and the exponent \(-x^2\) itself is a continuous function, their composition is also continuous everywhere. Therefore, the function \(h(x)\) is continuous on the closed interval \([-a, a].\)
02

Check Differentiability on (a, b)

To check for differentiability on the open interval \((-a, a),\) we first need to find the first derivative of the function \(h(x).\) Using the chain rule, we get: $$h'(x) = \frac{d}{dx} e^{-x^2} = -2x \cdot e^{-x^2}.$$ The derivative \(h'(x) = -2x \cdot e^{-x^2}\) is a product of a polynomial function and an exponential function, which are both continuous on their domains. Therefore, the derivative is also continuous, and the function is differentiable on the open interval \((-a, a).\)
03

Check if h(a) = h(b)

Our interval is \([-a, a].\) We must check if \(h(-a) = h(a).\) Since the exponent in \(h(x)\) is an even number, we get \(h(-a) = e^{-(-a)^2} = e^{-a^2}\) and \(h(a) = e^{-a^2}.\) Thus, \(h(-a) = h(a).\)
04

Apply Rolle's Theorem and Find Point(s) c

Since the function \(h(x) = e^{-x^2}\) is continuous on the closed interval \([-a, a],\) differentiable on the open interval \((-a, a),\) and \(h(-a) = h(a),\) Rolle's Theorem applies to this function and interval. Therefore, there exists at least one point c 鈭� \((-a, a)\) such that \(h'(c) = 0.\) To find the point(s) c, we set \(h'(x) = 0\) and solve for x: $$-2x \cdot e^{-x^2} = 0.$$ In this equation, \(e^{-x^2}\) is nonzero for all x. So, to make the equation true, the term \(-2x\) must be equal to 0, which gives us \(x = 0.\) Consequently, there is exactly one point guaranteed by Rolle's Theorem: \(c = 0.\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Calculus
Calculus is a branch of mathematics that studies how things change. It provides us tools to analyze the behavior of functions and their rates of change.
  • In calculus, we often study the properties of functions, such as continuity and differentiability.
  • Calculus is fundamental for understanding various scientific and engineering concepts, allowing us to solve problems involving changing systems.
A key part of calculus involves finding and understanding derivatives, which provides insights into how a function changes at any given point. In the context of Rolle's Theorem, calculus allows us to analyze if certain conditions make a function behave in a predictable way across an interval. Understanding these concepts helps in various applications, from physics to economics.
Continuity
In mathematics, a function is considered continuous if there are no breaks, holes, or gaps in its graph for a given interval. Continuity is a crucial aspect when applying mathematical theorems like Rolle's Theorem.
To determine if a function is continuous:
  • Check that the function is well-defined for every point in the interval.
  • Ensure that there are no abrupt changes in value within the interval.
For the function given as \[h(x) = e^{-x^2}\] on the interval \([-a, a]\), the behavior of exponential functions ensures continuity because they smoothly curve without breaks. This property lays the groundwork for differentiability and further application of calculus principles in analyzing critical points.
Differentiability
Differentiability refers to a function's ability to have a derivative, meaning it's smooth enough to have a tangent at every point in an interval. A function being differentiable on an interval is crucial for Rolle's Theorem.
To determine differentiability:
  • Find the derivative of the function. If the derivative is defined, the function is likely differentiable.
  • Check for smoothness, ensuring no sharp corners or cusps in the interval.
The function \[h(x) = e^{-x^2}\] is differentiable since the derivative \[-2x \, e^{-x^2}\] exists and is continuous within the interval \((-a, a)\). Smooth behavior here ensures that Rolle's Theorem can be applied, predicting the existence of at least one horizontal tangent line on the open interval.
First Derivative
The first derivative of a function provides the rate of change of the function's value concerning changes in the independent variable, usually x. It is a fundamental tool in calculus that helps identify critical points where the function might be increasing, decreasing, or have horizontal tangents.
For the function \[h(x) = e^{-x^2}\], its first derivative is \[h'(x) = -2x \, e^{-x^2}\]. It reflects how the function changes as x changes.
  • If the first derivative is zero at a point, the function has a horizontal tangent at that point.
  • The existence of such a point is central to Rolle's Theorem, which assures at least one such critical point exists if the theorem's conditions are met.
In the solution's context, solving the equation \[-2x \, e^{-x^2} = 0\] yields \[x = 0\], indicating a critical point where the first derivative is zero, perfectly aligning with the expectations set by Rolle's Theorem.

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