Problem 12 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

For each function: (a) Find all critical points on the specified interval. (b) Classify each critical point: Is it a local maximum, a local minimum, an absolute maximum, or an absolute minimum? (c) If the function attains an absolute maximum and/or minimum on the specified interval, what is the maximum and/or minimum value? \(f(x)=3 x^{4}-8 x^{3}+3\) on \((0,3)\)

Step-by-Step Solution

Answer

The critical point is at \(x=2\), it is a point of inflection. The function attains an absolute maximum value of 6 at \(x=3\) and an absolute minimum value of 3 at \(x=2\)

Step 1: Compute the first derivative of the function

The first derivative of the function \(f(x)=3 x^{4}-8 x^{3}+3\) is given by \[f'(x)=12x^{3}-24x^{2}\] using the power rule.

Step 2: Determine the critical points

The critical points are found by setting the derivative equal to zero. So, solve the equation \(f'(x)=12x^{3}-24x^{2}=0\). This simplifies to \(x^{2}(12x-24)=0\), giving the critical points at \(x=0\) and \(x=2\). Note that \(x=0\) is not in the interval (0, 3)

Step 3: Compute the second derivative of the function

The second derivative of the function is given by \[f''(x)=36x^{2}-48x\] using the power rule.

Step 4: Determine the nature of the critical point

Finding the nature of the critical point at \(x=2\) by evaluating \(f''(2)=36*2^{2}-48*2=0\). This result indicates that \(x=2\) is a point of inflection.

Step 5: Evaluate the function at the endpoints and the critical point

Find the values of the function at \(x=0, 2, 3\). These are \(f(0)=3\), \(f(2)=3\) and \(f(3)=6\) respectively.

Step 6: Determine the absolute maximum and/or minimum

Compare the function values at the critical point and the endpoint. The maximum value is the greatest of these values and the minimum is the least. So, \(f(3)=6\) is the maximum value and \(f(2)=f(0)=3\) is the minimum value.

Key Concepts

First Derivative Test

The First Derivative Test is a critical tool for understanding the behavior of a function at its critical points. Critical points occur where the first derivative is zero or undefined. These points can indicate potential local maxima, minima, or neither.

When a function's first derivative changes from positive to negative, it suggests a local maximum at that critical point. Conversely, if the derivative changes from negative to positive, we often have a local minimum. If there is no change in the sign of the derivative, it could imply a point of inflection or a saddle point.

For instance, in our example with the function f(x) = 3x^4 - 8x^3 + 3, the critical points were found after setting the first derivative f'(x) = 12x^3 - 24x^2 equal to zero. Critical point at x=2 does not change the sign of the derivative, indicating that it could be a point of inflection or a saddle point rather than a local maximum or minimum. To effectively apply the First Derivative Test, always remember to check the signs of the derivative before and after each critical point.

Second Derivative Test

The Second Derivative Test helps to further classify critical points, especially when the First Derivative Test is inconclusive. This test relies on the concavity of a function at a particular point, determined by the value of the second derivative.

If the second derivative at a critical point is positive, f''(x) > 0, then the function is concave up at that point, and you have a local minimum. If it's negative, f''(x) < 0, the function is concave down, indicating a local maximum. When the second derivative is zero, the test is inconclusive; the critical point could be a point of inflection.

In our example, we computed the second derivative f''(x) = 36x^2 - 48x and found that at the critical point x=2, the second derivative equals zero, which means the Second Derivative Test is inconclusive. It's worth noting that the critical point at x=0 isn't considered since it lies outside the specified interval of (0,3). The test supports the finding that the critical point at x=2 may be an inflection point.

Power Rule Differentiation

The Power Rule for Differentiation is a fundamental technique in calculus used to differentiate functions of the form f(x) = ax^n, where a is a constant, and n denotes the power or exponent. The rule states that the derivative of such a function is f'(x) = nax^(n-1).

Applying this rule simplifies the process of finding derivatives, which is essential for locating critical points and analyzing the behavior of functions. For the given function in our exercise, f(x) = 3x^4 - 8x^3 + 3, we applied the Power Rule to obtain the first derivative, f'(x) = 12x^3 - 24x^2 and the second derivative f''(x) = 36x^2 - 48x.

The use of the Power Rule made it possible to quickly find the derivatives necessary for applying the First and Second Derivative Tests to identify and classify the critical points of the function. Remember, when using the Power Rule, keep an eye on the sign and magnitude of the exponent; these will greatly influence the slope and concavity of the function at various points.