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

Question

In Problems 1 through 12: (a) Find all critical points. (b) Find \(f^{\prime \prime} .\) Use the second derivative, wherever possible, to determine which critical points are local maxima and which are local minima. If the second derivative test fails or is inapplicable, explain why and use an alternative method for classifying the critical point. $$ f(x)=4 x^{-1}+2 x^{2} $$

Step-by-Step Solution

Answer

The function \( f(x)=4 x^{-1}+2 x^{2} \) has two critical points: \( x=1 \) and \( x=-1 \). The point \( x=1 \) is a local minimum, while \( x=-1 \) is neither a local maximum nor a minimum.

Step 1: Calculate The First Derivative Of The Function

To find the critical points of the function, we first need to compute its derivative. The derivative of the function \( f(x)=4 x^{-1}+2 x^{2} \) is given by \( f'(x) = -4x^{-2} + 4x \) by applying the power rule.

Step 2: Find The Critical Points

The critical points are the solutions to the equation \( f'(x) = 0 \), and the values of x where \( f'(x) \) doesn't exist. Solving \( -4x^{-2} + 4x = 0 \) for \( x \), we get two critical points: \( x=1 \) and \( x=-1 \) . So \( x=1 \) and \( x=-1 \) are the critical points.

Step 3: Calculate The Second Derivative Of The Function

To use the second derivative test where possible, we need to compute the second derivative of the function. The second derivative of the function is given by \( f''(x) = 8x^{-3} + 4 \), again by applying the power rule.

Step 4: Classify The Critical Points

By inserting the critical points into the second derivative: the value of \( f''(1) \) is 12, which is positive, indicating that \( x=1 \) is a local minimum. On the other hand, \( f''(-1) \) doesn't exist, indicating that the second derivative test fails for \( x=-1 \). Therefore, we can use the first derivative test as an alternative. To the left of \( -1 \), \( f' \) is positive and to the right of \( -1 \), \( f' \) is also positive. Therefore, \( x=-1 \) is neither a local maximum nor a minimum.

Key Concepts

Critical Points

When studying calculus, one of the most important aspects is finding the points at which a function changes its behavior. These points are known as critical points. Critical points are found where the first derivative of a function either equals zero or does not exist.

• **Solutions to the equation**: To find the critical points, you solve the equation where the first derivative equals zero.
• **Derivative's non-existence**: Look also for points where the first derivative is undefined, as these can indicate sharp turns or cusps in the graph.

In our solution, we calculated the critical points of the function by solving the equation derived from the first derivative. We found two critical points, at \( x = 1 \) and \( x = -1 \). These points are where the function potentially has local maxima, minima, or other changes in its behavior.
Understanding critical points is fundamental because they are often where local maximum and minimum values occur, giving insight into the function's graph and overall behavior.

First Derivative

The first derivative of a function is a powerful tool in understanding how that function behaves. It shows us the rate of change of the function at any point, helping us predict whether the graph is going up or down.
Calculating the first derivative involves basic differentiation rules like the power rule. For example, the derivative of the function \( f(x) = 4x^{-1} + 2x^{2} \) is computed as \( f'(x) = -4x^{-2} + 4x \). This showcases how each term in the function behaves individually.

• **Significance of zero**: Critical points arise when this first derivative is zero because it indicates moments of no change, where the slope of the tangent line is horizontal.
• **Change in sign**: If the first derivative changes sign from positive to negative or vice versa, it confirms a peak or trough at that point in the function.

Using the first derivative is a direct method to identify critical points and understand the increase/decrease behavior of a function.

Second Derivative

Once you have the critical points, using the second derivative helps determine the nature of these critical points. The second derivative provides information on the concavity of the function.
Calculating the second derivative involves differentiating the first derivative. In our exercise, \( f''(x) = 8x^{-3} + 4 \) is obtained by applying the power rule again.

• **Concavity**: Positive second derivative indicates the graph is concave up (local minimum potential), while a negative value indicates concave down (local maximum potential).
• **Test application**: The second derivative test is applied to determine if a critical point is a local minimum or maximum. If the test fails, alternative methods must be used.

For instance, when we apply the second derivative to \( x = 1 \), the positive result confirms a local minimum, while at \( x = -1 \), the test fails due to non-existence, leading to further investigation.

Local Maxima and Minima

Determining the local maxima and minima of a function involves looking at the critical points and analyzing the second derivative's results.
Local maxima and minima are the largest and smallest values in the region around a point, akin to peaks and valleys on a graph.

• **Using second derivative results**: Positive results in the second derivative at a critical point suggest a local minimum, while negative results suggest a local maximum.
• **Alternative methods**: If the second derivative test fails, such as being undefined, use the first derivative test to check changes in the sign which can indicate rise or fall transitions.

In our case, at \( x = 1 \), the function has a local minimum based on the positive second derivative value. However, at \( x = -1 \), the test was inconclusive, and further examination was required, revealing neither a local maximum nor minimum. This holistic analysis is essential for confirming the specific behavior of different segments of the function's graph.