Problem 42 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, find the maximum value of \(\left|f^{\prime \prime}(x)\right|\) on the closed interval. (You will use this skill in Section \(12.4\) to estimate the error in the Trapezoidal Rule.) $$ f(x)=\frac{1}{x^{2}+1}, \quad[0,3] $$

Step-by-Step Solution

Answer

The maximum value of \( |f''(x)| \) on the interval [0,3] occurs at x = 0,1, and 3.

Step 1: Differentiation of the function

Differentiate the function to get \( f'(x) \). The first derivative is given by using the quotient rule for derivatives \( (u/v)' = (u'v-uv')/v^2 \) where \( u = 1 \) and \( v = x^{2}+1 \). The derivative is \( f'(x) = -2x/(x^{2}+1)^2 \).

Step 2: Second Differentiation of the function

Differentiate the first derivative to get the second derivative, \( f''(x) \). Apply the quotient rule again. Checking for points which make this expression 0, which are also called critical points.

Step 3: Find the critical points

Equating the second derivative to zero and solving for \( x \), we get three critical points: \( x = 0 \), and \( x = \pm 1 \). It is important to confirm that these points lie in the interval [0,3]. Note that we also consider the endpoints of the interval as potential locations of the maximum and minimum of \( |f''(x)| \)

Step 4: Function evaluation at critical points and interval endpoints

Evaluate \( |f''(x)| \) at the critical points \( x = 0 \), \( x = 1 \) and at the endpoints of the interval \( x = 0 \) and \( x = 3 \). Choose the highest value among all.

Key Concepts

Understanding the Maximum Value of the Second Derivative

To find the maximum value of the second derivative, we first differentiate the function twice. In this context, the function is \( f(x) = \frac{1}{x^2 + 1} \). Differentiating it once gives us \( f'(x) \), the first derivative. We then apply differentiation again to get \( f''(x) \), the second derivative.

Finding the maximum value of \( |f''(x)| \) helps us understand how rapidly the function's slope is changing across the interval \([0,3]\). This value is crucial when using numerical methods, like the Trapezoidal Rule, as it allows us to estimate potential errors. Remember, we're interested in the highest absolute value, not just the highest value, to cover both increases and decreases.

Identifying Critical Points

Critical points are where the derivative equals zero or is undefined. These points often correspond to local maxima or minima of a function.

For our function, the critical points are determined by setting the second derivative \( f''(x) \) to zero and solving for \( x \). In our example, these points include \( x = 0 \) and \( x = \pm 1 \). Only the points within the interval \([0,3]\) are relevant. Critical points help us identify where potential maximum or minimum values of \( |f''(x)| \) might occur.

Function Evaluation at Key Points

Once we have our critical points, we need to evaluate \( |f''(x)| \) at these points and at the endpoints of the interval, \( x = 0 \) and \( x = 3 \). This evaluation determines which point gives the maximum absolute value of the second derivative.

This step ensures we capture the behavior of the function across the entire specified interval. Evaluating at critical and endpoint locations provides a comprehensive view of where the greatest change occurs.

Applying the Quotient Rule

The quotient rule is essential when differentiating functions presented as a fraction. For a quotient \( \frac{u}{v} \), the rule is \( (\frac{u}{v})' = \frac{u'v - uv'}{v^2} \).

In our scenario, \( u = 1 \) and \( v = x^2 + 1 \). Using this rule helps in accurately finding \( f'(x) = -\frac{2x}{(x^2 + 1)^2} \), and subsequently \( f''(x) \), which is needed to pinpoint critical points and extreme values.

Understanding the quotient rule is crucial for handling more complex derivatives, particularly when dealing with rational functions.