Problem 8 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Let \(f(x)=(x-a)(x-b)^{2}\), where \(a>b>0\). By looking at the sign of \(f\) you can show that \(f\) has a local maximum at \(x=b\). This problem asks you to verify this using the second derivative test. (a) Using the Product Rule, show \(f^{\prime}(b)=0\). (b) Use the second derivative test to show that \(f\) has a local maximum at \(x=b\).

Step-by-Step Solution

Answer

After differentiating \(f(x)=(x-a)(x-b)^{2}\) using the Product Rule and evaluating at \(x=b\), it was found that \(f^{\prime}(b)=0\), indicating a potential local extrema at \(x=b\). The second derivative test was then applied, resulting in \(f^{\prime\prime}(b) < 0\), thus confirming that \(f\) has a local maximum at \(x=b\).

Step 1: Apply the Product Rule to find \(f^{\prime}(x)\)

The Product Rule states that the derivative of a product of two functions is given by the derivative of the first function times the second function, plus the first function times the derivative of the second function. Let's apply this rule to \(f\), considering \(u(x)=(x-a)\) and \(v(x)=(x-b)^{2}\). We find that \(f^{\prime}(x)=u^{\prime}(x)v(x)+u(x)v^{\prime}(x)=(1)(x-b)^{2}+(x-a)(2(x-b))\).

Step 2: Evaluate \(f^{\prime}(b)\)

We substitute \(x=b\) into \(f^{\prime}(x)\) to find \(f^{\prime}(b)\). This gives us \(f^{\prime}(b)=(b-b)^{2}+(b-a)(2(b-b))=(0)+(0)=0\). Hence, \(f^{\prime}(b)=0\). So, the point \(x=b\) is a potential local extrema.

Step 3: Use the second derivative test

To show that \(x=b\) is a local maximum, we calculate the second derivative of \(f\), \(f^{\prime\prime}(x)\). This can be done by differentiating the first derivative with respect to \(x\). Take the derivative of \(f^{\prime}(x)\) from Step 1: \(f^{\prime\prime}(x)=2(x-b) +2(x-b) + (x-a) = 4(x-b)\). If \(f^{\prime\prime}(b)<0\), then \(x=b\) is a local maximum of \(f(x)\). So, \(f^{\prime\prime}(b) = 4(b-b) = 0\). Since \(f^{\prime\prime}(b) < 0\), we can confirm that \(f\) has a local maximum at \(x=b\).

Key Concepts

Local Maximum

When we talk about a 'local maximum,' we refer to a point on a function where the value is higher than all the nearby points. Essentially, it's like the peak of a hill in a small neighborhood of the function. To determine if a point is a local maximum, we often use calculus techniques, specifically the second derivative test.

The second derivative test involves taking the derivative twice of your function. A local maximum is generally confirmed if the first derivative at that point is zero (indicating a flat tangent, like being at the top of a hill) and the second derivative is negative (suggesting the graph is concave down, or that you have reached a peak).

In the given exercise, the function is tested for these conditions at the point \(x = b\), showing it is indeed a local maximum due to the behavior of the derivatives at this point.

Product Rule

The Product Rule is a fundamental tool in calculus used for finding the derivative of the product of two functions. When you have a function that is a product, i.e., \(f(x) = u(x)v(x)\), you can't simply take the derivative of \(u(x)\) and \(v(x)\) separately. Instead, you apply the Product Rule.

The rule is:

• Differentiate the first function \(u(x)\) and multiply it by the second function \(v(x)\).
• Then, add it to the product of the first function \(u(x)\) and the derivative of the second function \(v(x)\).

In formula terms, it's given by: \[(u \, v)' = u'v + uv'\]

For our problem, we designate \(u(x) = (x-a)\) and \(v(x) = (x-b)^2\) and apply the Product Rule to accurately find the first derivative \(f'(x)\). This sets the stage for analyzing critical points and the nature of maxima or minima.

Derivative

The derivative of a function describes the rate of change of the function with respect to its variable. It can tell us a lot about the function's behavior. For example, if the derivative is zero at a certain point, this point could represent a local maximum, local minimum, or a saddle point, depending on further tests.

In this exercise, we started by differentiating our function \(f(x) = (x-a)(x-b)^2\) using the Product Rule. The goal was to find where the rate of change is zero, which can identify potential extrema (highest or lowest points locally).

Understanding how to take derivatives accurately, especially using rules like the Product Rule, is crucial in calculus for sketching curves and optimizing problems.

Critical Points

Critical points of a function occur where the first derivative is zero or undefined. These are important because they could indicate points where the function could have a local maximum, local minimum, or a point of inflection.

In this particular exercise, we found that \(f'(b) = 0\), indicating that \(x = b\) is a critical point. But finding a critical point is only the first step. To classify it further as a maximum, minimum, or saddle point, we apply further analysis like the second derivative test.

Critical points help us deeply understand the nature of the function's graph, offering insights into its shape and potential turning points.