Problem 9 - 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? \(f(x)=x^{5}-20 x+5\) on \([0,2]\)

Step-by-Step Solution

Answer

The function has a critical point at \(x = \sqrt{2}\) which is a local minimum. The function attains an absolute maximum value of 5 at \(x = 0\) and an absolute minimum value of -3 at \(x = 2\).

Step 1: Find the derivative

Find the derivative of \(f(x)=x^{5}-20 x+5\). This gives \(f'(x) = 5x^{4}-20\).

Step 2: Find the critical points

Critical points occur where the derivative is zero or undefined. The derivative is a polynomial and so it's defined everywhere. We set the derivative equal to zero to find critical points. So, solve the following equation: \(5x^{4}-20 = 0\). This simplifies to \(x^{4} = 4\), hence \(x = \sqrt[4]{4} = (2^{\frac{1}{2}})^{\frac{1}{2}}\) which means \(x = \sqrt{2}\). So, the only critical point in the interval \([0,2]\) is \(x = \sqrt{2}\).

Step 3: Classify the critical point

We classify the critical point by using the first derivative test. Into the first derivative, substitute a test point taken from each interval of the number line separated by the critical point. Here, intervals are \([0, \sqrt{2})\) and \((\sqrt{2}, 2]\). The two test points can be 1 and 2. Substituting these test points into \(f'(x)\):- for \(x = 1\), we get \(f'(1) = 5(1^{4})-20 = -15 < 0\) which indicates a decreasing function.- for \(x = 2\), we get \(f'(2) = 5(2^{4})-20 = 60 > 0\) which indicates an increasing function.So, \(x = \sqrt{2}\) is a local minimum.

Step 4: Find the absolute maxima or minima

To find the absolute maximum and minimum values of the function on the closed interval \([0,2]\), we have to evaluate the function at its critical point and at the endpoints of the interval. So, evaluate \(f(x)\) at \(x = 0\), \(x = \sqrt{2}\), and \(x = 2\). This gives \(f(0) = 5\), \(f(\sqrt{2}) = (\sqrt{2})^{5} - 20\sqrt{2} + 5\), \(f(2) = 32 - 40 + 5 = -3\). Hence, by comparison, the function attains its absolute maximum of 5 at x = 0 and its absolute minimum of -3 at x = 2.

Key Concepts

Critical Points

Critical points of a function are the values of the input or independent variable where the function's derivative is either zero or undefined. These are important because they potentially mark the locations where the function changes behavior. For the function in question, \(f(x) = x^5 - 20x + 5\), we first determine its derivative, giving us \(f'(x) = 5x^4 - 20\). Since the derivative is defined everywhere for real values, we find the critical points by setting \(f'(x) = 0\). Solving this, we are left with \(x^4 = 4\), which leads to the critical point \(x = \sqrt{2}\) within the interval \([0, 2]\).
Finding critical points is a key step in analyzing a function as they are the potential locations for local extrema.

Derivative Test

To classify the critical points further, we employ the derivative test. One commonly used technique involves the first derivative test. This test examines the sign changes of \(f'(x)\) in intervals around the critical point. For \(x = \sqrt{2}\), the derivative test involves examining the intervals \([0, \sqrt{2})\) and \((\sqrt{2}, 2]\). By substituting test points—such as \(1\) for the left interval and \(2\) for the right—into the derivative:

• For \(x = 1\), \(f'(1) = -15\), showing a decrease as it is negative.
• For \(x = 2\), \(f'(2) = 60\), showing an increase as it is positive.

This switch from negative to positive indicates that the critical point \(x = \sqrt{2}\) is a local minimum. This systematic methodical approach to testing gives confidence in identifying the nature of critical points.

Local Maximum and Minimum

A local maximum or minimum occurs at a critical point where the function shows differing behavior on either side. As identified using the derivative test, \(x = \sqrt{2}\) is a local minimum. This is because the function goes from decreasing to increasing at this point, leading it to dip down at \(x = \sqrt{2}\) forming a valley. Conversely, a local maximum would show a peak where the function transitions from increasing to decreasing. Local extrema are useful for understanding a function's shape, but aren't always the highest or lowest points globally, especially when analyzing over specific intervals.

Absolute Maximum and Minimum

The absolute maximum and minimum of a function over a closed interval are the highest and lowest values the function achieves on that interval, respectively. To find these absolute extrema, we not only check the critical points but also the endpoints of the interval. For the function \(f(x) = x^5 - 20x + 5\) on \([0, 2]\), evaluate the function at critical point \(x = \sqrt{2}\) and endpoints \(x = 0\) and \(x = 2\):

• \(f(0) = 5\)
• \(f(\sqrt{2}) = (\sqrt{2})^5 - 20\sqrt{2} + 5\)
• \(f(2) = -3\)

Comparing these values reveals the absolute maximum as \(5\) at \(x = 0\) and the absolute minimum as \(-3\) at \(x = 2\). Evaluating endpoints along with critical points ensures a comprehensive understanding of an interval's extrema.