Problem 27 - Essential Calculus

Question

Find all relative extrema. Use the Second Derivative Test where applicable. \(f(x)=\cos x-x, \quad[0,4 \pi]\)

Step-by-Step Solution

Answer

The relative extrema of \(f(x)=\cos(x) - x\) on the interval \([0, 4\pi]\) are minima at \(\pi\) and \(3\pi\).

Step 1: Derivative Computation

Start by finding the derivative of \(f(x)\): \(f'(x) = -\sin x - 1\).

Step 2: Find Critical Points

Set this equal to zero to find the critical points for extrema: \(-\sin x - 1 = 0\). Solving this gives: \(\sin x = -1\). In the interval \([0, 4\pi]\), this happens at \(\pi, 3\pi\), which are the critical points.

Step 3: Second Derivative Computation

To apply the second-derivative test, find the second derivative of \(f(x)\): \(f''(x) = -\cos x\).

Step 4: Applying Second Derivative Test

Substitute the critical points into the second derivative: \(f''(\pi) = -\cos(\pi) = 1\) and \(f''(3\pi) = -\cos(3\pi) = 1\). Since the second derivative at both points is greater than zero, both are relative minima.

Step 5: Checking Endpoints

Finally, we also have to consider the endpoints of the interval. \(f(0) = \cos(0) - 0 = 1\) and \(f(4\pi) = \cos(4\pi) - 4\pi = 1 - 4\pi\). Hence these are not relative extrema.

Key Concepts

Understanding the Derivative

The derivative of a function gives us crucial information about the rate of change of that function. In the context of our exercise, the function is \( f(x) = \cos x - x \). We find the derivative \( f'(x) = -\sin x - 1 \). The derivative tells us how the original function \( f(x) \) is changing as \( x \) changes. When the derivative is zero, the function is neither increasing nor decreasing, indicating a potential point of interest, which can be a peak, trough, or a turning point.

In simpler terms, when the function is at a relative high or low point, the slope of the tangent (i.e., the derivative) is zero. This is why we set \( f'(x) = 0 \) to find these points. For \( f(x) = \cos x - x \), this equation simplifies to \( -\sin x - 1 = 0 \), helping us locate values of \( x \) where possible relative extrema occur.

Second Derivative Test Explained

The Second Derivative Test is a powerful tool to determine whether a critical point is a minimum or maximum. Once you find the first derivative and the critical points, the second derivative \( f''(x) \) gives us insight into the curvature of \( f(x) \) at those points. For our function, the second derivative is \( f''(x) = -\cos x \).

By evaluating \( f''(x) \) at the critical points, you can decide the nature of these points:

• If \( f''(x) > 0 \), the function is concave up, suggesting a local minimum.
• If \( f''(x) < 0 \), the function is concave down, suggesting a local maximum.
• If \( f''(x) = 0 \), the test is inconclusive.

In this exercise, \( f''(\pi) = 1 \) and \( f''(3\pi) = 1 \), indicating both are relative minima since \(1 > 0\). The second derivative confirms the curvature, acting like a magnifying glass to specify the type of extremum at critical points.

Critical Points and Their Importance

Critical points are where the derivative of a function is zero or undefined. These points indicate potential locations for relative minima or maxima. In simple terms, these are points where the function might switch from increasing to decreasing or vice versa.

In the example \( f(x) = \cos x - x \), the critical points come from solving \( -\sin x - 1 = 0 \), giving us \( x = \pi, 3\pi \) within the interval \([0, 4\pi]\).

These points help us identify behavior changes in \( f(x) \). While all critical points are points of interest, not all are points of relative extrema, which is why other tests like the second derivative test are necessary to make further distinctions.

Identifying Relative Extrema

Relative extrema refer to the peaks and troughs of a function on a given interval. They represent the highest or lowest points relative to their immediate surrounding points. For functions, relative minima and maxima are relevant to understanding how the function behaves.

In the exercise, after identifying the critical points \( \pi \) and \( 3\pi \) and applying the second derivative test, both were found to be relative minima because \( f''(x) > 0 \) at these points.

It’s important to test the endpoints of the interval here: \( f(0) = 1 \) and \( f(4\pi) = 1 - 4\pi \). However, neither endpoint qualifies as a relative extremum because they aren't the highest or lowest within the interval relative to their neighbors. Thus, the relative extrema within \([0, 4\pi]\) are at \( x = \pi \) and \( x = 3\pi \). This distinction allows us to focus on parts of the function where extreme values are most compelling.