Problem 4 - 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. $$ f(x)=x^{5}-5 x $$

Step-by-Step Solution

Answer

Critical points are \(x = 1\) and \(x = -1\). Using the Second Derivative Test, \(x = 1\) is a local minimum and \(x = -1\) is a local maximum.

Step 1: Finding the derivative

To find the critical points of the function \(f(x)=x^{5}-5x\), first calculate the derivative of \(f(x)\) using the power rule, which gives \(f'(x)=5x^{4}-5\).

Step 2: Finding the critical points

The critical points are found by setting the derivative equal to zero and solving for \(x\). So, setting \(5x^{4}-5 = 0\) simplifies to \(x^{4} = 1\), which gives the critical points as \(x = 1\) and \(x = -1\).

Step 3: Finding the second derivative

To determine local maxima and minima, find the second derivative of the function, which is \(f''(x)=20x^{3}\).

Step 4: Applying the Second Derivative Test

To apply the second derivative test, plug the critical points into the second derivative. \(f''(1)=20\) and \(f''(-1)=-20\). Since \(f''(1) > 0\), the point at \(x = 1\) is a local minimum. Similarly, as \(f''(-1) < 0\), the point at \(x = -1\) is a local maximum, according to the Second Derivative Test.

Key Concepts

Finding Critical Points

Understanding how to find critical points in a function is a fundamental aspect of calculus. Critical points are where the function's derivative is zero or undefined, indicating a potential local maximum, local minimum, or saddle point (where the function has a horizontal tangent but is neither a max nor a min).

To locate these critical points, as in the exercise with the function f(x) = x^5 - 5x, we begin by finding the derivative, using rules of differentiation like the power rule. We then set the derivative f'(x) equal to zero and solve for x. The solutions to this equation are the critical values—unless the derivative doesn't exist at some points, which we must also consider. In the given exercise, the critical points were found at x = 1 and x = -1, by solving 5x^4 - 5 = 0.

It's important to remember that not all critical points will lead to a maximum or minimum; they simply indicate a place where the slope of the tangent is zero. To determine the nature of these points, we could use other methods such as the First Derivative Test or the Second Derivative Test.

Second Derivative Test

The Second Derivative Test is a convenient way to determine whether a critical point is a local maximum, local minimum, or neither. When the second derivative at a critical point is positive (f''(x) > 0), the function is concave up, suggesting a local minimum. Conversely, if it's negative (f''(x) < 0), the function is concave down, and we have a local maximum.

In our problem, the second derivative is f''(x) = 20x^3. Substituting the critical values x = 1 and x = -1 into f''(x), we obtain f''(1) = 20 and f''(-1) = -20. Therefore, based on the Second Derivative Test, x = 1 is a local minimum and x = -1 is a local maximum. This test can sometimes fail, for instance, if the second derivative is zero or undefined at the critical point. In such cases, we would resort to alternative methods, like the First Derivative Test, to classify the critical points.

Power Rule Differentiation

The power rule is a vital tool in differentiation, immensely simplifying the process of finding derivatives, especially for polynomials. According to this rule, if you have a function f(x) = x^n, where n is any real number, the derivative of that function is f'(x) = nx^(n-1).

In the context of our exercise, we applied the power rule to differentiate the function f(x) = x^5 - 5x. We treated each term separately, differentiating x^5 to get 5x^4 and -5x to get -5. The combined result was the derivative f'(x) = 5x^4 - 5.

This rule is straightforward and practical for finding derivatives quickly, making it easier to calculate rates of change, slopes of tangent lines, and—as we have seen—finding critical points. It's a cornerstone of calculus that facilitates the analysis of functions by breaking down complex operations into manageable steps.