Problem 180 - Calculus Volume 1

Question

Determine whether the Mean Value Theorem applies for the functions over the given interval \([a, b] .\) Justify your answer. $$ y=x \sin (\pi x) \text { over }[0,2] $$

Step-by-Step Solution

Answer

The Mean Value Theorem applies to the function on the interval \([0, 2]\).

Step 1: Check Continuity

The first requirement for the Mean Value Theorem is that the function must be continuous on the closed interval \([0, 2]\). The function \(y = x \sin(\pi x)\) is a product of \(x\) and \(\sin(\pi x)\). Both \(x\) and \(\sin(\pi x)\) are continuous for all real numbers, hence \(y = x \sin(\pi x)\) is continuous on \([0, 2]\).

Step 2: Check Differentiability

The second requirement is that the function must be differentiable on the open interval \((0, 2)\). To check differentiability, compute the derivative of \(y\). Use the product rule, \(u'v + uv'\), where \(u(x) = x\) and \(v(x) = \sin(\pi x)\). Therefore, \(u' = 1\) and \(v' = \pi \cos(\pi x)\). The derivative \(y' = 1\cdot \sin(\pi x) + x \cdot (\pi \cos(\pi x)) = \sin(\pi x) + \pi x \cos(\pi x)\). Since sine and cosine functions are differentiable for all real numbers, and \(x\) is polynomial (thus differentiable), \(y = x \sin(\pi x)\) is differentiable on \((0, 2)\).

Step 3: Conclusion

Both continuity on \([0, 2]\) and differentiability on \((0, 2)\) are satisfied for \(y = x \sin(\pi x)\). Thus, the Mean Value Theorem applies to this function over the interval \([0, 2]\).

Key Concepts

Understanding Continuity

Continuity is a fundamental concept in calculus, and it's essential when discussing the Mean Value Theorem. In simple terms, a function is continuous over an interval if there are no breaks, jumps, or holes in its graph. For a more formal definition: a function \( f(x) \) is continuous at a point \( c \) if the limit of \( f(x) \) as \( x \) approaches \( c \) from both directions is equal to \( f(c) \).
To apply the Mean Value Theorem, the function needs to be continuous over the closed interval \([a, b]\), including the endpoints.
When checking if the function \( y = x \sin(\pi x) \) is continuous over \([0, 2]\), we consider each part of the product:

• The function \( x \) is continuous for all real numbers because it's a polynomial.
• The function \( \sin(\pi x) \) is continuous for all real numbers since sine functions do not have any discontinuities.

Thus, their product \( y = x \sin(\pi x) \) remains continuous over the interval \([0, 2]\). This property ensures that we can apply the Mean Value Theorem, provided other conditions also hold.

Essentials of Differentiability

Differentiability is another crucial concept when applying the Mean Value Theorem. A function \( f(x) \) is differentiable at a point \( c \) if it has a defined derivative at that point. Essentially, this means the function's graph at \( c \) can be smoothly approximated by its tangent.
To apply the Mean Value Theorem, the function must be differentiable on the open interval \((a, b)\). Differentiability on \( (0, 2) \) means there are no sharp corners or cusps.
For the function \( y = x \sin(\pi x) \), we find the derivative using the product rule. The product rule helps us differentiate functions that are the product of two simpler functions. It states:

• If you have a function \( y = u \cdot v \), then its derivative is \( y' = u'v + uv' \).
• Here, \( u(x) = x \) and \( v(x) = \sin(\pi x) \).
• The derivatives are \( u' = 1 \) and \( v' = \pi \cos(\pi x) \).

So, the derivative \( y' \) becomes \( \sin(\pi x) + \pi x \cos(\pi x) \). Both sine and cosine are differentiable everywhere, and multiplying by \( x \) or \( \pi x \) doesn't disrupt differentiability. Thus, \( y \) is differentiable on \((0, 2)\). This fulfills another requirement for the Mean Value Theorem.

Applying the Product Rule

The product rule is a technique used in calculus to differentiate functions that are expressed as a product of two other functions. This rule is vital for handling complex expressions like \( y = x \sin(\pi x) \), where neither factor by itself is constant.
The product rule is expressed by the formula:

• \( y = u \cdot v \) implies \( y' = u' v + u v' \).

Let’s break down the differentiation of the function \( y = x \sin(\pi x) \):

• Identify \( u(x) = x \) and \( v(x) = \sin(\pi x) \).
• Calculate the derivatives: \( u' = 1 \) and \( v' = \pi \cos(\pi x) \).
• Substitute these into the product rule formula: \( y' = 1 \cdot \sin(\pi x) + x \cdot \pi \cos(\pi x) \).

So, the derivative \( y' \) is \( \sin(\pi x) + \pi x \cos(\pi x) \). This product rule ensures that the function is differentiable where needed, providing us with a powerful tool for calculus problems and specifically verifying conditions like those required for the Mean Value Theorem. Understanding this rule can greatly simplify working with complex functions in calculus.