Problem 14 - 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 \([-3,0)\)

Step-by-Step Solution

Answer

The critical points on [-3, 0] are \(-\sqrt{3}\) (local maximum and absolute minimum) and 0 (neither maximum nor minimum). The absolute maximum is at -3.

Step 1: Deriving the function

First, let's find the derivative of \(f(x)\). Remember the power rule for derivatives, \(\frac{d}{dx}x^n=nx^{n-1}\), and the rule for the derivative of a reciprocal, \(\frac{d}{dx}1/x=-1/x^2\). So, \(f'(x) = \frac{d}{dx}(\frac{x^{3}}{3})+ \frac{d}{dx}(2x) + \frac{d}{dx}(\frac{3}{x}) = x^{2}+2-3/x^2.\)

Step 2: Find critical points

The critical points are where the derivative is equal to zero or undefined. So,1. Set \(f'(x)=0: x^{2}+2-3/x^2=0\). Multiply by \(x^2\) to clear the fraction -> \(x^4+2x^2-3=0\).2. The solutions to this equation are \(x= -\sqrt{3}, x=0\).3. Let's check where \(f'(x)\) is undefined: Never, as the domain of \(f'(x)\) is all real numbers except 0, but 0 is a critical point as seen above. Result: The critical points in the interval [-3,0] are \(-\sqrt{3}\) and 0.

Step 3: Classifying critical points

Now, let's find the second derivative and classify the critical points:- \(f''(x)=2x+6/x^3\). At \(x=-\sqrt{3}\), \(f''(-\sqrt{3})<0\) so it's a local maximum.- At \(x=0\), \(f''(x)\) is undefined, so we can't use the second derivative test. But observe that \(f'(x)>0\) for \(x<0\) and \(f'(x)<0\) for \(x>0\). Hence, \(x=0\) is neither a maximum nor a minimum. So the local maximum is at \(x=-\sqrt{3}\).

Step 4: Absolute maximum and minimum

We need to compare the values of \(f(x)\) at the critical points and the ends of the interval. But the function is undefined at \(x=0\), so we only need to compare \(f(-3)\) and \(f(-\sqrt{3})\). Since \(f(-3)>f(-\sqrt{3})\), \(-3\) is the absolute maximum and \(-\sqrt{3}\) is the absolute minimum.

Key Concepts

Derivative of Functions

Understanding the derivative of a function is a cornerstone of calculus. In essence, the derivative represents the rate at which a function is changing at any given point. For polynomial functions, we apply the power rule, which for a term like \(ax^n\) gives us a derivative of \(anx^{n-1}\). When encountering terms like \(1/x\), which is \(x^{-1}\), the derivative is \(-1/x^2\) due to applying the same power rule.

When we derived the function \(f(x)=\frac{x^3}{3}+2x+\frac{3}{x}\), we applied these rules. The derivative \(f'(x)\), which signifies the slope of the tangent to the curve represented by \(f(x)\), is found to be \(x^2+2-3/x^2\). Knowing the derivative is key to finding critical points, which are points where the slope of the tangent is zero or the derivative is undefined, indicating potential peaks, valleys, or flat regions on the curve of \(f(x)\).

Classifying Critical Points

Identifying the critical points is one thing but classifying them—determining whether they're local maxima, local minima, or saddle points—calls for further investigation. In the exercise, we begin by setting \(f'(x)\) to zero to find potential extremum points. With \(f'(x)\) identified as \(x^2+2-3/x^2\), we saw that the critical points lie where \(f'(x) = 0\). After some algebraic manipulation, we found the critical points to be \(-\r\tsqrt{3}\) and 0 on the interval \([-3,0)\).

To classify these points, we examined the second derivative, \(f''(x)\). This practice is grounded in the second derivative test, where a positive \(f''(x)\) at a critical point implies a local minimum, and a negative implies a local maximum. If \(f''(x)\) is zero or undefined, the test is inconclusive. For \(x = -\r\tsqrt{3}\), a negative second derivative suggested a local maximum. However, at \(x = 0\), since the second derivative was undefined, we could not use it to classify the critical point.

Absolute Maximum and Minimum

Critical points can also be used to determine where a function reaches its highest or lowest values on a given interval, which are classified as absolute maximum and minimum values. This is different from the local perspective as it considers the entire range of interest. The process of finding these absolute values involves comparing function values at critical points and the endpoints of the interval.

Returning to our problem with \(f(x)\), we compared \(f(-3)\) and \(f(-\r\tsqrt{3})\) since these are the boundary and the critical point within the interval \([-3,0)\). Remember that the function’s behavior at \(x=0\) is not included, as specified by the open interval. By evaluating \(f(x)\) at \(x=-3\) and \(x=-\r\tsqrt{3}\), we concluded that \(-3\) is the location of the absolute maximum for \(f(x)\) on \([-3,0)\). With these evaluations, we confirmed the presence and the nature of absolute extremes within the specified domain, a key step in understanding the function's overall behavior on that interval.