Problem 11 - 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.$$ f(x)=\frac{x^{5}}{5}-x^{4}+\frac{4}{3} x^{3}+2 $$

Step-by-Step Solution

Answer

The critical points of the function are x = 0 and x = 2. The point x = 2 is a local maximum. The second derivative test fails at x = 0.

Step 1: Identify the Function

We are given function \(f(x) = \frac{x^{5}}{5} - x^{4} + \frac{4}{3}x^{3} + 2\)

Step 2: Find the First Derivative

By applying the power rule for differentiation, the first derivative of the function f'(x) is obtained by differentiating each term in the function with respect to x. So, \(f'(x) = x^{4} - 4x^{3} + 4x^{2}\)

Step 3: Find Critical Points

The critical points are obtained by setting the first derivative equal to zero, and solving for \(x\). So, we solve the equation \(x^{4} - 4x^{3} + 4x^{2} = 0\). Factoring it, we get \(x^{2}(x - 2)² = 0\). Therefore, the critical points are \(x = 0, 2\)

Step 4: Find the Second derivative

By applying the power rule for differentiation to the first derivative, the second derivative f''(x) is obtained. So, \(f''(x) = 4x^{3}-12x^{2}+8x\)

Step 5: Classify the Critical Points

Apply the second derivative test to classify the critical points found in step 3. This involves plugging the critical points into the second derivative and analyzing the sign. For x = 0, f''(0) = 0, second derivative test fails. For x = 2, f''(2) = 4(2)³ - 12(2)² + 8(2) = -16, which is less than 0, hence x = 2 is a local maximum. For the point where the second derivative test failed, we can use the first derivative test or close-value test.

Key Concepts

First Derivative Test

The First Derivative Test is a method used to determine if a function has a local maximum, a local minimum, or neither at its critical points. A critical point occurs where the first derivative of a function is zero or undefined, indicating a potential change in direction for the function's graph.

To use the First Derivative Test:

• Calculate the first derivative of the function.
• Find critical points by setting the first derivative equal to zero. Solve for these points.
• Test these critical points by observing the sign of the derivative before and after them.

If the derivative changes from positive to negative at a critical point, the function has a local maximum there. If it changes from negative to positive, the function has a local minimum. If there is no sign change, and the derivative doesn't become zero, the critical point is neither a maximum nor a minimum.

Second Derivative Test

The Second Derivative Test is a mathematical tool used to determine the concavity of a function and to classify critical points even further. This test can provide information about whether these critical points are local maxima, local minima, or points of inflection, based on the concavity.

Here’s how to apply it:

• Take the second derivative of the function.
• Substitute the critical points into the second derivative.
• Analyze the sign of the results:

If the second derivative at a critical point is positive, the function is concave up and has a local minimum at that point. If it is negative, the function is concave down and has a local maximum. However, if the second derivative is zero, the test is inconclusive. An alternative method, like the First Derivative Test, must then be used to analyze the critical point.

Differentiation

Differentiation is the mathematical process of finding the derivative of a function. The derivative represents the rate of change of the function's value with respect to changes in the input value. It is a crucial concept for understanding motion, growth, and change within the realm of calculus.

To differentiate a polynomial function like the one given:

• Apply the power rule, which states that for a term ax^n, the derivative is nax^{n-1}.
• Calculate the derivative for each term separately.
• Sum these derivatives to get the first derivative of the entire function.

In the original exercise solution, the first derivative of the provided function is calculated as:\[ f'(x) = x^{4} - 4x^{3} + 4x^{2} \] This derivative is then used to find the critical points by setting it equal to zero and solving for "x".

Local Maximum

A local maximum is a point where a function attains a peak value within a small interval. The value of the function at this point is greater than the values in its immediate neighborhood.

To determine a local maximum:

• Identify the critical points using the first derivative; these are potential candidates.
• Perform the First Derivative Test to see if there is a change from positive to negative around the critical point, indicating a peak or local maximum.
• If the Second Derivative Test is applicable, substitute critical points into the second derivative. If it yields a negative result, it confirms a local maximum due to downward concavity.

For example, in the exercise solution, the critical point at \(x = 2\) is confirmed as a local maximum because the function's second derivative is negative at that point: \[ f''(2) = -16 \] This indicates a concave down shape, signifying a local maximum.

Local Minimum

A local minimum is a point at which a function reaches its lowest value in a nearby interval. The value at a local minimum is less than the values close to it.

The process to find a local minimum includes:

• Locate the critical points using differentiation.
• Utilize the First Derivative Test to evaluate if there is a change from negative to positive at these points, indicating a valley or local minimum.
• If the Second Derivative Test is used, input the critical points into the second derivative. A positive result confirms a local minimum due to upward concavity.

In the provided exercise, the Second Derivative Test was not applicable for some critical points due to a zero result. In such cases, you should rely on an alternative method like the First Derivative Test to conclude if the point is a local minimum.