Problem 13 - 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? \(f(x)=\frac{x^{3}}{3}+2 x+\frac{3}{x}\) on its natural domain. Why is \(x=0\) not a critical point?

Step-by-Step Solution

Answer

The critical points are \(x = -1, 1\), with \(x=-1\) being a local (and absolute) maximum and \(x=1\) being a local (and absolute) minimum. The function does not have a critical point at \(x=0\) because \(x=0\) is not within its domain.

Step 1: Finding the Derivative

To find the critical points of the function \(f(x) = \frac{x^{3}}{3}+2x+\frac{3}{x}\), first find the derivative \(f'(x)\). Use the rule \((u^n)' = nu^{(n-1)}\) for the first term, the identity rule for the second, and the rule \((\frac{1}{v})' = -\frac{v'}{v^2}\) for the last term. This gives: \[f'(x) = x^2 + 2-\frac{3}{x^2}\]

Step 2: Finding Critical Points

Critical points occur when \(f'(x) = 0\) or \(f'(x)\) is undefined. In our case, \(f'(x)\) is undefined at \(x=0\), but since \(x=0\) is not in the domain of the original function, it's not a critical point. Therefore, set \(f'(x) = 0\) and solve for \(x\): \[x^2 + 2 = \frac{3}{x^2} \ x^4 + 2x^2 - 3 = 0 \ (x^2-1)(x^2+3)=0 \ x=±1\] So, the critical points are \(x = -1, 1\)

Step 3: Classify Each Critical Point

To determine whether the critical points are local maxima or minima, use the sign of the second derivative \(f''(x)\). The second derivative is: \[f''(x) = 2x + \frac{6}{x^3}.\] Now, plug the critical points into the second derivative. If \(f''(x) < 0\), then \(x\) is a local maximum; if \(f''(x) > 0\), then \(x\) is a local minimum. So, for \(x=1\), \(f''(x) = 2 + \frac{6}{1} = 8 > 0\), which means \(x=1\) is a local minimum, and for \(x=-1\), \(f''(-1) = -2 + \frac{6}{-1} =-8 < 0\), which means \(x=-1\) is a local maximum.

Step 4: Finding the Absolute Maximum and/or Minimum

Since the function doesn't have any endpoints (it spans \(-\infty\) to \(\infty\)), to find the absolute maximum and minimum values, it's enough to evaluate the function at its critical points. Substituting \(x = -1\) into the function, we get \(f(-1) = -\frac{1}{3} -2 -3 < 0\), and for \(x = 1\), \(f(1) = \frac{1}{3} + 2 + 3 > 0\). So, \(x=-1\) gives the absolute maximum, \(x=1\) gives the absolute minimum.

Step 5: Why \(x=0\) is not a critical point?

As previously stated, \(x=0\) is outside the function's domain, therefore, while the derivative is undefined at \(x=0\), it is not considered a critical point.

Key Concepts

Finding Critical Points in Calculus

When studying calculus, one of the essentials is understanding how to find critical points of a function. Critical points are where the function's derivative is either zero or undefined, and these points often indicate where the function experiences its peaks and troughs.

To pinpoint critical points, firstly, calculate the derivative of the function using the appropriate rules of differentiation. For the function in question, \(f(x) = \frac{x^{3}}{3}+2x+\frac{3}{x}\), we employ power rules and the reciprocal rule to find \(f'(x)\).

Once you have the derivative, analyze where \(f'(x) = 0\) or where \(f'(x)\) is not defined. It's key to remember that points where the function is not defined, like \(x=0\) in this case, are not part of the domain and thus not considered critical points. After solving \(f'(x) = 0\), you determine the critical points, \(x = -1\) and \(x = 1\) for our example.

Classifying Critical Points

The next step after finding critical points is to determine their nature. Critical points can be classified as local maxima, minima, or saddle points based on the behavior of the function around these points. To classify them, you can use the second derivative test.

By taking the second derivative \(f''(x)\), and substituting the critical points into it, the function’s concavity at those points is revealed. If \(f''(x) > 0\), the function is concave up, indicating a local minimum. Conversely, if \(f''(x) < 0\), the function is concave down, and you're at a local maximum. In our example, \(f''(1) = 8\) suggests a local minimum at \(x = 1\), and \(f''(-1) = -8\) indicates a local maximum at \(x = -1\).

Absolute Maximum and Minimum

Lastly, establishing the presence of absolute maximum and minimum values is crucial. These are the highest and lowest values that a function achieves in its domain. Unlike local maxima or minima, absolute values are not about the function's shape but its highest and lowest outputs.

To find these values when the function is continuous on a closed interval, evaluate the function at critical points and at the boundaries of its domain. If the domain is not bounded, like with \(f(x)\), then, in many cases, just considering the critical points suffices. For the provided function, you evaluate \(f(-1)\) and \(f(1)\) to uncover that \(x=-1\) yields the absolute maximum and \(x=1\) the absolute minimum since the function spans negative to positive infinity and these are the only critical points.

In summary, mastering these concepts—finding and classifying critical points and understanding absolute extrema—is essential for analyzing functions in calculus and important across various applications in mathematics and the physical sciences.