Problem 25 - Calculus A New Horizon

Question

Analyze the trigonometric function \(f\) over the specified interval, stating where \(f\) is increasing. decreasing, concave up, and concave down, and stating the \(x\) coordinates of all inflection points. Confirm that your results are consistent with the graph of \(f\) generated with a graphing utility. $$f(x)=\cos x ;[0,2 \pi]$$

Step-by-Step Solution

Answer

Decreasing on \((0, \pi)\), increasing on \((\pi, 2\pi)\); concave down on \((0, \pi)\), concave up on \((\pi, 2\pi)\); inflection points at \(x = \frac{\pi}{2}, \frac{3\pi}{2}\).

Step 1: Finding the First Derivative

To analyze where the function \(f(x) = \cos x\) is increasing or decreasing, we first find the derivative. The derivative is \(f'(x) = -\sin x\).

Step 2: Identify Critical Points

Set the first derivative equal to zero to find critical points: \(-\sin x = 0\). Solving this within the interval \([0, 2\pi]\) gives us \(x = 0, \pi, 2\pi\).

Step 3: Test the Intervals for Increase/Decrease

Evaluate the sign of \(f'(x)\) (\(-\sin x\)) in the intervals \((0, \pi)\) and \((\pi, 2\pi)\). - In the interval \((0, \pi)\), \(-\sin x\) is negative, so \(f(x)\) is decreasing. - In the interval \((\pi, 2\pi)\), \(-\sin x\) is positive, so \(f(x)\) is increasing.

Step 4: Finding the Second Derivative

To determine concavity and find inflection points, compute the second derivative of \(f(x)\): \(f''(x) = -\cos x\).

Step 5: Determine Concavity

Evaluate the sign of \(-\cos x\). - From \((0, \pi)\), \(-\cos x\) is negative, so \(f(x)\) is concave down. - From \((\pi, 2\pi)\), \(-\cos x\) is positive, so \(f(x)\) is concave up.

Step 6: Identify Inflection Points

Inflection points occur when \(f''(x) \) changes sign. From \(-\cos x = 0\), we have \(x = \frac{\pi}{2}, \frac{3\pi}{2}\). These points separate concave up from concave down.

Key Concepts

Derivatives

The derivative is an essential concept when analyzing a function's behavior. It tells us how the function is changing at any point and helps us understand whether the function is increasing or decreasing. For the function \(f(x) = \cos x\), we find the first derivative to be \(f'(x) = -\sin x\). This derivative indicates the rate of change of \(f(x)\) at each point in the interval \([0, 2\pi]\).

By setting the derivative equal to zero, \(-\sin x = 0\), we can find the critical points, which are potential turning points of the function. In this case, \(x = 0, \pi, 2\pi\) are our critical points within the interval.

Understanding derivatives is crucial because it provides the foundation for more advanced concepts like concavity and inflection points.

Critical Points

Critical points in calculus are where the function's derivative equals zero or where the derivative is undefined. They are significant because they help identify potential maximums, minimums, or points of inflection. In the context of the trigonometric function \(f(x) = \cos x\) over the interval \([0, 2\pi]\), the first derivative \(f'(x) = -\sin x\) is used to find critical points.

Setting \(-\sin x = 0\) results in the critical points \(x = 0, \pi, 2\pi\). These points divide the interval into smaller sections, which can be tested to determine whether the function is increasing or decreasing in each section. Here, between \(0\) and \(\pi\), the function decreases, and between \(\pi\) and \(2\pi\), the function increases.

Critical points are merely indicators of potential changes in the function's behavior. Additional analysis with the second derivative can indicate true changes in concavity.

Concavity

Concavity refers to the curvature of a function. When a function is concave up, it means it is curving upwards, like a cup. Conversely, when it is concave down, it curves downwards, like an upside-down cup. The sign of the second derivative determines concavity. For \(f(x) = \cos x\), we compute the second derivative \(f''(x) = -\cos x\).

By analyzing \(-\cos x\) over the interval \([0, 2\pi]\):

• In the interval \((0, \pi)\), \(-\cos x\) is negative. This indicates that \(f(x)\) is concave down.
• In the interval \((\pi, 2\pi)\), \(-\cos x\) is positive. This signifies that \(f(x)\) is concave up.

Understanding concavity is vital because it provides insights into the nature of the function's graph, which can explain changes in direction and behavior more deeply.

Inflection Points

Inflection points are where the concavity of a function changes. At these points, the second derivative either changes from positive to negative or vice versa. For the function \(f(x) = \cos x\), we look at the second derivative, \(f''(x) = -\cos x\), to identify these points.

By setting \(-\cos x = 0\), we find potential inflection points at \(x = \frac{\pi}{2}, \frac{3\pi}{2}\). These are the points where the function transitions from being concave down to concave up, or vice versa.

Inflection points are critical in understanding the "shape" of the function's graph, as they highlight changes in the direction of the curvature, which can't be determined from just the first derivative.