Problem 33 - Calculus Early Transcendentals

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. \(f(x)=\sin x-\cos x ;[-\pi, \pi]\)

Step-by-Step Solution

Answer

The function is increasing on \((-\frac{3\pi}{4}, \pi]\) and decreasing on \([-\pi, -\frac{3\pi}{4})\). It is concave up on \((-\frac{\pi}{4}, \frac{3\pi}{4})\) with inflection points at \(x = -\frac{\pi}{4},\frac{3\pi}{4}\).

Step 1: Find the Derivative

To determine where the function is increasing or decreasing, we first find the derivative of the function. Given the function \(f(x) = \sin x - \cos x\), the derivative, using the rules \(\frac{d}{dx}(\sin x) = \cos x\) and \(\frac{d}{dx}(\cos x) = -\sin x\), is \(f'(x) = \cos x + \sin x\).

Step 2: Determine Critical Points

We find the critical points by setting \(f'(x) = 0\). So, \(\cos x + \sin x = 0\). Solving this equation, we get \(\sin x = -\cos x\). Dividing both sides by \(\cos x\), we have \(\tan x = -1\). Thus, the solutions within \([-\pi, \pi]\) are \(x = -\frac{3\pi}{4}\) and \(x = \frac{\pi}{4}\).

Step 3: Analyze Intervals for Increasing/Decreasing

We analyze intervals determined by the critical points: \([-\pi, -\frac{3\pi}{4})\), \((-\frac{3\pi}{4}, \frac{\pi}{4})\), and \((\frac{\pi}{4}, \pi]\). Calculate \(f'(x)\) at test points in each interval: \(f'(-\pi) = -1\), \(f'(-\frac{\pi}{2}) = \frac{1}{\sqrt{2}} > 0\), and \(f'(\frac{\pi}{2}) = \frac{1}{\sqrt{2}} > 0\) respectively. It shows that the function is decreasing on \([-\pi, -\frac{3\pi}{4})\) and increasing on \((-\frac{3\pi}{4}, \pi]\).

Step 4: Find the Second Derivative

To determine concavity, we find the second derivative of \(f(x)\). Since \(f'(x) = \cos x + \sin x\), the second derivative is \(f''(x) = -\sin x + \cos x\).

Step 5: Determine Inflection Points

We find inflection points by setting \(f''(x) = 0\), giving \(-\sin x + \cos x = 0\). Solving this, we get \(\sin x = \cos x\). Dividing by \(\cos x\), we have \(\tan x = 1\). Thus, the solutions within \([-\pi, \pi]\) are \(x = -\frac{\pi}{4}\) and \(x = \frac{3\pi}{4}\).

Step 6: Analyze Intervals for Concavity

We use the second derivative to check the sign at test points: \(f''(-\frac{\pi}{2}) = -1 < 0\), \(f''(0) = 1 > 0\), and \(f''(\pi) = -1 < 0\). This implies concave down on \((-\pi, -\frac{\pi}{4})\) and \((\frac{3\pi}{4}, \pi)\) and concave up on \((-\frac{\pi}{4}, \frac{3\pi}{4})\).

Key Concepts

Critical Points

Critical points are essential in understanding a function's behavior. These occur where the derivative equals zero or is undefined. To find the critical points for the function \(f(x) = \sin x - \cos x\) within the interval \([-\pi, \pi]\), we start with the derivative, \(f'(x) = \cos x + \sin x\). We set it to zero for critical points: \(\cos x + \sin x = 0\). Solving it leads to \(\tan x = -1\), identifying points \(x = -\frac{3\pi}{4}\) and \(x = \frac{\pi}{4}\). - Critical points help determine intervals where the function changes from increasing to decreasing or vice versa.- They are usually checked further for whether those intervals are indeed increasing or decreasing. - These points serve as potential locations of local maxima or minima.

Concavity

Concavity tells how the curve bends. It's determined by the second derivative of the function. For \(f(x) = \sin x - \cos x\), the first derivative \(f'(x) = \cos x + \sin x\) leads us to the second derivative \(f''(x) = -\sin x + \cos x\). By setting \(f''(x) = 0\), we find potential inflection points: \(-\sin x + \cos x = 0\), or \(\tan x = 1\). Solving within \([-\pi, \pi]\) gives \(x = -\frac{\pi}{4}\) and \(x = \frac{3\pi}{4}\).- Concave up indicates the graph is like a cup (curving upwards), while concave down is like a cap (curving downwards).- Using test points, we analyze intervals: With \(f''(-\frac{\pi}{2}) = -1 < 0\), the function is concave down on \((-\pi, -\frac{\pi}{4})\) and \((\frac{3\pi}{4}, \pi)\).- Alternatively, \(f''(0) = 1 > 0\) implies the function is concave up on \((-\frac{\pi}{4}, \frac{3\pi}{4})\).

Increasing and Decreasing Functions

Understanding where a function is increasing or decreasing is key in graph analysis. A function increases where its first derivative \(f'(x)\) is positive, and decreases where \(f'(x)\) is negative. With \(f'(x) = \cos x + \sin x\) for the function \(f(x) = \sin x - \cos x\) on \([-\pi, \pi]\), we analyze around our critical points \(x = -\frac{3\pi}{4}\) and \(x = \frac{\pi}{4}\).- Test points in intervals - At \(-\pi\), \(f'(-\pi) = -1\). Here, \(f(x)\) is decreasing. - At \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\), \(f'\) yields positive values \(\frac{1}{\sqrt{2}} > 0\). Thus, \(f(x)\) is increasing. - Thus, we conclude \(f(x)\) decreases on \([-\pi, -\frac{3\pi}{4})\) and increases on \((-\frac{3\pi}{4}, \pi]\).- These findings are crucial in understanding the overall graph shape and behavior of the trigonometric function across the interval.