Problem 50 - Single Variable Calculus

Question

In Exercises \(41-60\), find the absolute maximum and absolute minimum values, if any, of the function. $$ f(x)=2 x+\frac{1}{x} \text { on }[-1,3] $$

Step-by-Step Solution

Answer

The function \(f(x) = 2x + \frac{1}{x}\) has an absolute maximum value of \(6\frac{1}{3}\) at x = 3, and an absolute minimum value of -3 at x = -1 on the interval \([-1, 3]\).

Step 1: Check continuity on the interval [-1, 3]

The given function \[f(x) = 2x + \frac{1}{x}\] is continuous on the interval \([-1,3]\), except for x = 0. Since the interval given is [-1,3], the function is continuous on this interval, and we can move to the next step.

Step 2: Calculate the first derivative and find the critical points

To find the critical points of the function, we first need to calculate the first derivative. Using the power rule of differentiation, we have: \[f'(x) = \frac{d}{dx}(2x) + \frac{d}{dx}(\frac{1}{x})\] \[f'(x) = 2 - \frac{1}{x^2}\] Now, we find the critical points by setting the first derivative equal to zero and solving for x: \[2 - \frac{1}{x^2} = 0\] \[\frac{1}{x^2} = 2\] \[x^2 = \frac{1}{2}\] \[x = \pm \frac{1}{\sqrt{2}}\] We found two critical points, \(x = \frac{1}{\sqrt{2}}\) and \(x = -\frac{1}{\sqrt{2}}\). Since \(-\frac{1}{\sqrt{2}}\) lies in the interval \([-1, 3]\), we will use this critical point for our analysis.

Step 3: Evaluate the function at the critical points and end points

Now we will evaluate the function at the critical point \(-\frac{1}{\sqrt{2}}\) and the interval endpoints, which are -1 and 3. \[f(-\frac{1}{\sqrt{2}}) = 2(-\frac{1}{\sqrt{2}}) + \frac{1}{(-\frac{1}{\sqrt{2}})} = -\sqrt{2} + \sqrt{2} = 0\] \[f(-1) = 2(-1) + \frac{1}{-1} = -2 - 1 = -3\] \[f(3) = 2(3) + \frac{1}{3} = 6 + \frac{1}{3} = 6\frac{1}{3}\]

Step 4: Compare the function values to find the absolute maximum and minimum

We have evaluated the function at the critical point and the interval endpoints, and now we will compare them to find the absolute maximum and minimum. - The function has an absolute minimum value of -3 at x = -1, as this is the smallest value. - The function has an absolute maximum value of \(6\frac{1}{3}\) at x = 3, as this is the largest value. So, the absolute maximum value of the function is \(6\frac{1}{3}\) at x = 3, and the absolute minimum value is -3 at x = -1.

Key Concepts

Critical Points

Understanding where a function reaches its highest highs and lowest lows demands that we find its 'critical points.' These are places where the function does not have a derivative (points where the function isn't smooth) or where the derivative is zero (the slopes of tangents are flat).

Imagine trekking across a mountain range; critical points are akin to the peaks and valleys - places where the trail either flattens out or is non-differentiable, like a sharp cliff edge.

For the function \(f(x) = 2x + \frac{1}{x}\), the critical points occur where the first derivative \(f'(x)\) equals zero. After calculating \(f'(x) = 2 - \frac{1}{x^2}\), we solve \(2 - \frac{1}{x^2} = 0\) to find the critical points at \(x = \frac{1}{\sqrt{2}}\) and \(x = -\frac{1}{\sqrt{2}}\). In this context, since we're only considering the interval [-1, 3], the critical point of interest is \(x = -\frac{1}{\sqrt{2}}\).

First Derivative Test

The first derivative test is a useful tool for analyzing the highs and lows—peaks and troughs—of a function's journey. Essentially, it helps us classify our critical points. It works like this: if you pass from negative derivative (downhill slope) to positive derivative (uphill slope), voila, you've found a valley, otherwise known as a local minimum. Conversely, going from positive (uphill) to negative (downhill) signals a peak, a local maximum.

Here's the catch though: if the derivative does not change signs, you might just be traversing a flat plateau, neither gaining nor losing elevation, which indicates a possible point of inflection. For the function at hand, after finding our critical points, we then want to check the sign of the derivative before and after these points to categorize them correctly.

Continuity of a Function

Continuity is the mathematical equivalent of a smooth road. A function is continuous if you can draw it without lifting your pen: no gaps, no jumps, no abrupt turns. That's how you can tell your function won't suddenly teleport or vanish into thin air. In our exercise, \(f(x) = 2x + \frac{1}{x}\) is examined for this type of behavior on the interval [-1,3].

This function cruises along smoothly except for a point of drama: \(x = 0\), where things go haywire since dividing by zero is an absolute no-no in mathematics. It's like a bridge out on our road. However, since our interval is [-1, 3], and the function is between -1 and 3, we breathe a sigh of relief that our path is clear.

Closed Interval Method

When navigating the terrain of a function on a closed interval – think of it as a bounded stretch of landscape – there's a systematic way to scout for the function's highest and lowest points. This process is called the 'Closed Interval Method,' and it's like conducting a thorough search from end to end, ensuring no stone is left unturned.

To apply this method, start by ensuring the function is continuous on the entire interval, just so we know there are no missing bridges on our path. Then, find and evaluate the function's critical points within that interval. Don't forget about the endpoints, though – they could be hiding the function's absolute champions for maximum or minimum values.

In the given exercise, after checking the function at \(x = -\frac{1}{\sqrt{2}}\), we also evaluate the endpoints x = -1 and x = 3, akin to scouring the edge of the forests. These checks help us conclude that the absolute minimum value is -3 at x = -1, and the absolute maximum value is \(6\frac{1}{3}\) at x = 3. With the closed interval method, you're sure not to miss the forest for the trees when seeking those extreme values.