Problem 41 - Single Variable Calculus

Question

Find the relative extrema, if any, of the function. Use the Second Derivative Test, if applicable. $$ f(t)=2 t+\frac{1}{t} $$

Step-by-Step Solution

Answer

The function \(f(t) = 2t + \frac{1}{t}\) has a relative minimum at \(t = \frac{1}{\sqrt{2}}\) with \(f\left(\frac{1}{\sqrt{2}}\right) = 2\frac{1}{\sqrt{2}} + \sqrt{2}\), and a relative maximum at \(t = -\frac{1}{\sqrt{2}}\) with \(f\left(-\frac{1}{\sqrt{2}}\right) = 2\left(-\frac{1}{\sqrt{2}}\right) - \sqrt{2}\).

Step 1: Find the first derivative

To find the first derivative of the function, we'll use the power rule and sum/difference rule for differentiation. The first derivative is: $$ f'(t) = \frac{d}{dt} \left(2t + \frac{1}{t}\right) = 2 + \frac{d}{dt} \left(t^{-1}\right) = 2 - t^{-2} $$

Step 2: Find critical points

Now, we have to find the critical points by setting the first derivative equal to zero: $$ f'(t) = 0 $$ Substitute the expression of \(f'(t)\) we found in Step 1: $$ 2 - t^{-2} = 0 $$ Now, solve this equation for \(t\): $$ t^{-2} = 2 \\ t^{2} = \frac{1}{2} \\ t = \pm \frac{1}{\sqrt{2}} $$ So, we have two critical points: \(t_1 = \frac{1}{\sqrt{2}}\) and \(t_2 = -\frac{1}{\sqrt{2}}\).

Step 3: Find the second derivative

Next, we will find the second derivative of the function by differentiating the first derivative \(f'(t)\) with respect to \(t\): $$ f''(t) = \frac{d}{dt}\left(2 - t^{-2}\right) = 2t^{-3} $$

Step 4: Plug the critical points

Now we will plug the critical points found in Step 2 into the second derivative: $$ f''\left(\frac{1}{\sqrt{2}}\right) = 2\left(\frac{1}{\sqrt{2}}\right)^{-3} = 2\sqrt{2} $$ $$ f''\left(-\frac{1}{\sqrt{2}}\right) = 2\left(-\frac{1}{\sqrt{2}}\right)^{-3} = -2\sqrt{2} $$

Step 5: Analyze the results

Using the results from Step 4, we can determine the relative extrema: - \(f''\left(\frac{1}{\sqrt{2}}\right) = 2\sqrt{2} > 0\), so there is a relative minimum at \(t = \frac{1}{\sqrt{2}}\) - \(f''\left(-\frac{1}{\sqrt{2}}\right) = -2\sqrt{2} < 0\), so there is a relative maximum at \(t = -\frac{1}{\sqrt{2}}\) Therefore, the relative extrema of the function \(f(t) = 2t + \frac{1}{t}\) are: - Relative minimum at \(t = \frac{1}{\sqrt{2}}\) with \(f\left(\frac{1}{\sqrt{2}}\right) = 2\frac{1}{\sqrt{2}} + \sqrt{2}\) - Relative maximum at \(t = -\frac{1}{\sqrt{2}}\) with \(f\left(-\frac{1}{\sqrt{2}}\right) = 2\left(-\frac{1}{\sqrt{2}}\right) - \sqrt{2}\)

Key Concepts

First Derivative

Understanding the first derivative is crucial when analyzing the behavior of functions. It represents the rate of change of the function or, in simpler terms, its slope at a particular point. When we're given a function like \(f(t) = 2t + \frac{1}{t}\), finding the first derivative, denoted as \(f'(t)\), helps us map out where the function is increasing or decreasing.

By applying differentiation rules, as seen in the exercise, we find that \(f'(t) = 2 - t^{-2}\). At the points where \(f'(t) = 0\), the function's slope is zero, and these points are candidates for finding relative extrema, also known as local highs and lows. It’s like reaching the top or bottom of a hill when hiking—the path momentarily flattens.

Critical Points

Critical points occur where the first derivative is zero or undefined. These are the spots where the function could potentially have a peak or a trough—places where the graph levels off or has a horizontal tangent. In the process of finding these points for \(f(t) = 2t + \frac{1}{t}\), we set \(f'(t) = 2 - t^{-2}\) to zero and solved for \(t\), yielding \(t_1 = \frac{1}{\sqrt{2}}\) and \(t_2 = -\frac{1}{\sqrt{2}}\).

It's important to note that not all critical points will be extrema. That's why further analysis, often involving the second derivative, is needed to determine the actual nature of these points.

Second Derivative

While the first derivative gives us the slope of a function, the second derivative, denoted as \(f''(t)\), tells us about the curvature or concavity of the function. It's a measure of how the slope itself is changing. For our function, we found that \(f''(t) = 2t^{-3}\). A positive second derivative means the function is curving upwards, suggesting a concave up shape like a cup, while a negative second derivative indicates a concave down or a frown shape.

Knowing the second derivative's sign at the critical points gives insight into the type of extrema—whether it's a minimum or maximum. This is the foundation of the Second Derivative Test, which is our next topic of discussion.

Second Derivative Test

The Second Derivative Test is a handy tool that helps confirm whether a critical point is a local maximum or minimum. If you plug a critical point into the second derivative and get a positive number, like we did at \(t = \frac{1}{\sqrt{2}}\), that implies there's a relative minimum at that point. Conversely, a negative outcome, as seen at \(t = -\frac{1}{\sqrt{2}}\), indicates a relative maximum.

This test turns the abstract process of analyzing function behavior into a numbers game—positive or negative, minimum or maximum. It's very useful, but remember it only works if the second derivative exists and isn’t zero at the critical points; otherwise, we have to use other methods such as the First Derivative Test or examining the function's graph.