Problem 11 - 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 \([-1,1]\)

Step-by-Step Solution

Answer

The function \(f(x)=3 x^{4}-8 x^{3}+3\) has one critical point on the interval [-1,1] which is at \(x = 0\), and this point is a local maximum. The function attains an absolute maximum of 12 at \(x=-1\) and an absolute minimum of -2 at \(x=1\).

Step 1: Find the Derivative

The first step is to calculate the derivative of the function. We use the power rule of differentiation, which states that the derivative of \(x^n\) is \(n * x^{n-1}\). When differentiating \(f(x)=3 x^{4}-8 x^{3}+3\), the result is \(f'(x) = 12x^{3}-24 x^{2}\).

Step 2: Find the Critical Points

We set the derivative equal to zero to find the critical points. Thus, we solve \(12x^{3}-24 x^{2} = 0\). Factoring out \(12\), we find \(12(x^{2}(x - 2))=0\), and since the product between three numbers equals zero, at least one of them must be zero. This gives us critical points \(x = 0\) and \(x = 2\), but since \(2\) is not in the interval, we disregard it.

Step 3: Classify the Critical Points

From the first derivative test, if \(f'(x)\) changes sign at our point, we have a local extrema. The derivative to the left of \(0\) is: \(f'(-0.5)=12(-0.5)^3-24(-0.5)^2 = 3\), so it's positive. On the right, we have \(f'(0.5)=12(0.5)^3-24(0.5)^2 = -3\), which is negative. Hence, \(x = 0\) is a local maximum.

Step 4: Find Absolute Maximum and Minimum

To find absolute extrema on the interval, we evaluate the endpoints and all critical points. We have one point - \(0\). Therefore, we find the values of \(f(x)\) at these three points: \(f(x=-1) = 12, f(x=0) = 3, f(x=1) = -2\). Thus, the function has its absolute maximum \(12\) at \(x=-1\) and its absolute minimum \(-2\) at \(x=1\)

Key Concepts

Power Rule of Differentiation

The power rule of differentiation is an essential tool in calculus, especially useful when dealing with polynomial functions. This handy rule simplifies the process of finding the derivative of functions that are composed of powers of x.

To apply the power rule, simply take an exponent of any term, multiply by the coefficient, and subtract one from the exponent, resulting in the formula: if you have a term \(x^n\), its derivative is \(n \cdot x^{n-1}\). For example, when differentiating \(f(x) = 3x^4 - 8x^3 + 3\), you can apply the power rule to each term:

• The derivative of \(3x^4\) is \(4 \cdot 3x^{3} = 12x^{3}\).
• The derivative of \(-8x^3\) is \(3 \cdot (-8)x^{2} = -24x^{2}\).
• The constant \(3\) disappears upon differentiation, as constants have a derivative of zero.

By combining these results, the resulting derivative is \(f'(x) = 12x^3 - 24x^2\). This method streamlines the process and allows you to quickly compute derivatives.

First Derivative Test

The first derivative test is a critical tool to determine whether a critical point is a local maximum, local minimum, or neither. Once you've found the critical points by setting the derivative of the function to zero, you'll need to analyze how the sign of the first derivative changes around those points.

Here's how it works:

• First, compute the derivative of the function.
• Identify where the derivative equals zero or does not exist; these are your critical points.
• Based on the behavior of the first derivative around these points, determine the nature of each point:
• If \(f'(x)\) changes from positive to negative, the point is a local maximum.
• If \(f'(x)\) changes from negative to positive, the point is a local minimum.
• If there is no change, the point may be a saddle point or not a local extremum.

For the function \(f(x)=3 x^{4}-8 x^{3}+3\), at \(x=0\) we observe the sign of \(f'(x)\) changes from positive to negative across this critical point, indicating it is a local maximum.

Absolute Extrema

Absolute extrema are the highest or lowest points over a function's defined domain, not just around a localized area. Finding absolute extrema is crucial for understanding the overall behavior of functions, especially when considering optimization problems.

Here's the approach to identify absolute extrema within a specific interval such as \([-1, 1]\):

• Determine the derivative \(f'(x)\) and find critical points within the interval.
• Evaluate the function at the critical points and endpoints of the interval.
• The largest value you find will be the absolute maximum, and the smallest will be the absolute minimum on the interval.

For \(f(x)=3 x^{4}-8 x^{3}+3\) evaluated on \([-1, 1]\), we find:

• At \(x=-1\), \(f(x)\) is 12.
• At \(x=0\), \(f(x)\) is 3.
• At \(x=1\), \(f(x)\) is -2.

The absolute maximum value of 12 occurs at \(x=-1\), and the absolute minimum value of -2 occurs at \(x=1\). Recognizing these important points gives insight into the range of the function within the interval.

Local Extrema

Local extrema refer to points on a function where the function reaches a peak (local maximum) or a valley (local minimum) relative to the points directly surrounding it. Unlike absolute extrema, which consider the entire interval, local extrema focus only on small surrounding neighborhoods.

To spot local extrema, you'll need to determine where the first derivative equals zero and apply the first derivative test, assessing sign changes in \(f'(x)\) around these critical points:

• If \(f'(x)\) changes from positive to negative, the point is classified as a local maximum.
• If \(f'(x)\) changes from negative to positive, the point becomes a local minimum.

In our example function \(f(x) = 3x^4 - 8x^3 + 3\), we've previously determined \(x=0\) is a local maximum, given that \(f'(x)\) transitions from positive to negative as \(x\) goes through zero. Thus, understanding the nature of these points helps elucidate how the function behaves locally within small intervals. This knowledge aids in comprehending both the peaks and valleys on a graph.