Question

Show that the function \(y=x+1 / x\) (with \(x \neq 0\) ) has two relative extrema, one a maximum and the other a minimum. Is the "minimum" larger or smaller than the "maximum"? How is this paradoxical result possible?

Short answer

The minimum (2) is larger than the maximum (-2). This occurs because they are relative extrema, not absolute.

Step-by-step solution

Find the derivative

First, let's find the derivative of the function \( y = x + \frac{1}{x} \). Apply the derivative rules: the derivative of \( x \) is 1, and the derivative of \( \frac{1}{x} \) is \( -\frac{1}{x^2} \). Therefore, \( y' = 1 - \frac{1}{x^2} \).

Set derivative to zero

To find the critical points, set the derivative equal to zero: \[ 1 - \frac{1}{x^2} = 0 \].Solving this gives \( \frac{1}{x^2} = 1 \), or \( x^2 = 1 \). This implies \( x = 1 \) or \( x = -1 \).

Second derivative test

Find the second derivative to determine concavity:\[ y'' = \frac{d}{dx} \left(1 - \frac{1}{x^2}\right) \ = \frac{2}{x^3} \].Evaluate the second derivative at the critical points:1. At \( x = 1 \), \( y'' = \frac{2}{1} = 2 \) (positive: indicates a local minimum).2. At \( x = -1 \), \( y'' = \frac{2}{-1} = -2 \) (negative: indicates a local maximum).

Determine function values at critical points

Calculate \( y \) at the critical points to compare their values:- At \( x = 1 \), \( y = 1 + \frac{1}{1} = 2 \).- At \( x = -1 \), \( y = -1 + \frac{1}{-1} = -2 \).

Compare maximum and minimum

The local maximum value \( -2 \) at \( x = -1 \) is smaller than the local minimum value \( 2 \) at \( x = 1 \). This contrast comes from the nature of local extrema, which only refers to relative, not absolute, heights of neighboring values.

Key concepts

Derivative

The derivative essentially measures the rate at which a function changes. In calculus, it's like a speedometer for any mathematical function, telling you how fast or slow its value is tweaking as the input (usually represented by \(x\)) changes. To find the derivative of a function, you'll utilize rules such as the power rule, product rule, and quotient rule. Let's keep it simple by looking at the function given in the exercise, \(y = x + \frac{1}{x}\). - The derivative of \(x\) is straightforward: it’s 1.- For \(\frac{1}{x}\), you can use the power rule: the derivative is \(-\frac{1}{x^2}\).Putting it all together, the derivative \(y'\) becomes:\[ y' = 1 - \frac{1}{x^2} \]This expression is what you'll use to analyze how the function is changing at each point along its curve.

Critical Points

Critical points are like the crossroads on the road of a function where interesting things happen! These points occur where the derivative equals zero or is undefined, signaling potential peaks, troughs, or flat spots in the graph.For the function in the exercise, \(y = x + \frac{1}{x}\), finding critical points involves solving the equation from the derivative:\[ 1 - \frac{1}{x^2} = 0 \]This simplifies to \(x^2 = 1\), giving us \(x = 1\) and \(x = -1\). These values are your critical points, where either a local maximum or a local minimum can pop up. So always start by setting the derivative to zero to carve out these crucial highlights in your analysis.

Second Derivative Test

Once you've identified the critical points, it's time to determine the nature of these points—are they peaks (maxima), valleys (minima), or neither? This is where the second derivative test becomes a trusty tool.Compute the second derivative from \(y' = 1 - \frac{1}{x^2}\):\[ y'' = \frac{2}{x^3} \]Now, you simply plug in the critical points into this second derivative:- At \(x = 1\), \( y'' = 2 \), which is positive. This indicates the function curves upwards here, confirming a local minimum.- At \(x = -1\), \( y'' = -2 \), a negative value, revealing that the function curves downwards, affirming a local maximum.These signs help you depict the shape of the graph around each critical point without needing to sketch the whole function.

Local Extrema

Local extrema refer to the highest or lowest points in a specific neighborhood of a graph. In other words, these are the function's peaks or pits when viewed through a "local" lens rather than globally.For our function \(y = x + \frac{1}{x}\), we've calculated:- At \(x = 1\), the local minimum yields \(y = 2\).- At \(x = -1\), the local maximum gives \(y = -2\).Interestingly, in this scenario, the local maximum's actual value \(-2\) is less than the local minimum's value \(2\). How can that be possible?

• Local extrema are relative to their immediate surroundings, not necessarily to the whole function.
• In this case, around \(x = -1\), \(y\) is higher than at points very close to \(x = -1\), not necessarily higher than every other point on the graph.

This concept explains why the local maximum can be lower in height than the local minimum and that’s the beautiful paradox of local extrema.