Chapter 3: Problem 31
In Problems \(31-36,\) sketch the graph of the given function \(f\) in the region \((-\pi, \pi),\) unless otherwise indicated, labeling all extrema (local and global) and the inflection points and showing any asymptotes. Be sure to make use of \(f^{\prime}\) and \(f^{\prime \prime}\). \(f(x)=\cos x-\sin x\)
Short Answer
Step by step solution
Find the Derivative
Solve for Critical Points
Determine Concavity and Inflection Points
Evaluate Function at Critical Points
Sketch the Graph
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Critical Points
To find these critical points for \(f(x) = \cos x - \sin x\), we commence by differentiating the function:
- First derivative: \(f'(x) = -\sin x - \cos x\)
- \(\tan x = -1\)
- Critical points: \(x = -\frac{3\pi}{4}\) and \(x = \frac{\pi}{4}\) within the interval \((-\pi, \pi)\).
Analyzing critical points gives us valuable insights into where the graph changes direction, helping us understand the broader shape of the graph.
Inflection Points
By differentiating \(f'(x) = -\sin x - \cos x\) again, we obtain:
- Second derivative: \(f''(x) = -\cos x + \sin x\)
- \(\tan x = 1\)
- Inflection points: \(x = -\frac{\pi}{4}\) and \(x = \frac{3\pi}{4}\) within \((-\pi, \pi)\).
Function Derivatives
For our function, \(f(x) = \cos x - \sin x\):
- First derivative \(f'(x) = -\sin x - \cos x\) shows where the slope is zero, indicating critical points.
- Second derivative \(f''(x) = -\cos x + \sin x\) helps in finding inflection points by showing where the curvature changes.
By mastering derivatives, one gains a powerful toolset to analyze and graph complex trigonometric functions effortlessly.