Problem 39 - Essential Calculus

Question

Find all relative extrema. Use the Second Derivative Test where applicable. \(f(x)=\operatorname{arcsec} x-x\)

Step-by-Step Solution

Answer

The critical points are \(x = -1, 1\). The result of the Second Derivative Test will determine whether these points represent relative maxima or minima.

Step 1: Differentiate the Function

The first step is to differentiate the function \(f(x)= arcsec x - x\). The derivative of \(arcsec x\) is \(1/(|x|\sqrt{x^2 -1})\) and the derivative of \(x\) is 1. Thus, the first derivative \(f'(x)= \frac{1}{|x|\sqrt{x^2 -1}} -1\)

Step 2: Find the Critical Points

Critical points occur where the first derivative equals zero or does not exist. Solve for \(x\) in the equation \(f'(x) = 0\), by setting \(\frac{1}{|x|\sqrt{x^2 -1}} -1 = 0\). Solving this gives \(x = -1, 1\). Next, determine where \(f'(x)\) does not exist. Since the denominator can't be zero, we find \(x\) in the equation \(|x|\sqrt{x^2-1} = 0\), which has no real solutions. Therefore, the critical points are \(x = -1, 1\).

Step 3: Apply Second Derivative Test

Take the second derivative of \(f(x)\), namely \(f''(x)\). Then use Second Derivative Test, i.e., plug each of critical points into the second derivative. If the second derivative at a point is positive, then \(f(x)\) has a relative minimum there. If it's negative, \(f(x)\) has a relative maximum there. If it's zero, the test is inconclusive.

Key Concepts

Second Derivative Test

The Second Derivative Test helps determine if a function has a relative maximum, minimum, or if the test is inconclusive at critical points. After finding the critical points, you take the second derivative, denoted as \(f''(x)\). Here's how you apply the test:

• If \(f''(x) > 0\) at a critical point, the function has a relative minimum there.
• If \(f''(x) < 0\) at a critical point, the function has a relative maximum there.
• If \(f''(x) = 0\), the test is inconclusive. You may want to use other methods to analyze the behavior of the function.

This test is handy because it can save time by providing a quick indication of the nature of critical points. However, it relies on the existence of a non-zero second derivative. If \(f''(x)\) is zero, the behavior of the function near the critical point will need further investigation.

Critical Points

Critical points are points on the graph of a function where the derivative is zero or undefined. These are important because they signal potential locations for local maxima, minima, or saddle points. Here's how you find them:

• First, find the first derivative of the function, \(f'(x)\).
• Set \(f'(x) = 0\) and solve for \(x\) to find possible critical points.
• Identify any points where \(f'(x)\) does not exist.

For example, in the function \(f(x) = \operatorname{arcsec} x - x\), solving \(f'(x) = 0\) led us to the critical points \(x = -1\) and \(x = 1\). It's crucial to check both where the derivative is zero and undefined, as critical behaviors of the function can occur in either case.

Derivative of Arcsecant

The derivative of the arcsecant function can be a bit challenging to grasp at first. The function is defined as \(\operatorname{arcsec}(x)\), and its derivative is given by \(\frac{1}{|x|\sqrt{x^2-1}}\). This may seem complex, but here's a simple breakdown:

• The absolute value \(|x|\) ensures the expression is defined for both positive and negative values where arcsecant is defined.
• The square root expression \(\sqrt{x^2-1}\) comes from the trigonometric identity involving secant and helps account for the range of arcsecant.

When you're differentiating a function like \(f(x) = \operatorname{arcsec} x - x\), you use this rule. By differentiating \(\operatorname{arcsec} x\), you get \(\frac{1}{|x|\sqrt{x^2-1}}\) and then subtract the derivative of \(x\), which is 1. This gives the first derivative as \(f'(x) = \frac{1}{|x|\sqrt{x^2-1}} - 1\). Remembering the derivative of arcsecant is key for finding critical points and applying the Second Derivative Test.