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Chapter 3: Problem 33

Find the critical points of the function in the interval \([0,2 \pi] .\) Determine if each critical point is a relative maximum, a relative minimum, or neither. Use the Second-Derivative Test, when possible. Determine the points of inflection in the interval \([0,2 \pi]\). Then sketch the graph on the interval \([0,2 \pi]\) : $$ \int(x)=\sin x+\cos x $$

Short Answer

Expert verified
Critical points at \(\frac{\pi}{4}, \frac{5\pi}{4}\); \[x = \frac{\pi}{4}\] is a relative maximum; \[x = \frac{5\pi}{4}\] is a relative minimum; Inflection points at \(\frac{3\pi}{4}, \frac{7\pi}{4}\).

Step by step solution

01

- Find the first derivative

To find the critical points of the function, we first need to find its first derivative. The function given is \(f(x) = \sin x + \cos x\). Use the derivative rules for sine and cosine functions: \[f'(x) = \cos x - \sin x\].
02

- Set the first derivative to zero

Set the first derivative equal to zero to find the critical points:\[\cos x - \sin x = 0\]. This simplifies to \[\cos x = \sin x\]. To solve this equation, divide each term by \(\cos x\):\[1 = \tan x\], giving \[x = \frac{\pi}{4}, \frac{5\pi}{4}\] within the interval \([0, 2\pi]\).
03

- Verify the critical points

Verify the critical points fall within the interval \([0, 2\pi]\). The critical points are \[x = \frac{\pi}{4}, \frac{5\pi}{4}\] which are both within the given interval.
04

- Find the second derivative

To determine whether each critical point is a relative maximum, minimum, or neither, find the second derivative of the function: \[f''(x) = -\sin x - \cos x\].
05

- Apply the Second-Derivative Test

Evaluate the second derivative at the critical points:\[f''\left(\frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) - \cos\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2} < 0\]. This indicates \(x = \frac{\pi}{4}\) is a relative maximum. Similarly, \[f''\left(\frac{5\pi}{4}\right) = -\sin\left(\frac{5\pi}{4}\right) - \cos\left(\frac{5\pi}{4}\right) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} > 0\]. This indicates \(x = \frac{5\pi}{4}\) is a relative minimum.
06

- Determine points of inflection

A point of inflection is where the second derivative changes sign. Find when \[f''(x) = -\sin x - \cos x\] equals zero: \[-\sin x - \cos x = 0\] which simplifies to \[\sin x = -\cos x\]. This occurs at \[x = \frac{3\pi}{4}, \frac{7\pi}{4}\] to be within \([0, 2\pi]\).
07

- Sketch the graph

Using the information gathered - relative maximum at \(x = \frac{\pi}{4}\), relative minimum at \(x = \frac{5\pi}{4}\), and inflection points at \(x = \frac{3\pi}{4}\) and \(x = \frac{7\pi}{4}\) - sketch the function \(f(x) = \sin x + \cos x\) over the interval \([0, 2\pi]\). The critical points, maximums and minimums, and inflection points will guide the curvature of the graph.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

first derivative
To find critical points of a function, our first step is to determine the first derivative. The first derivative tells us the rate of change of the function. For a given function \( f(x) = \sin x + \cos x \), use the standard rules for the derivatives of sine and cosine: \( f'(x) = \cos x - \sin x \). This new function \( f'(x) \) provides critical information regarding the slopes of the function. By setting this first derivative equal to zero, \( \cos x - \sin x = 0 \), we can find where the original function has a slope of zero - these points are our critical points.
second derivative test
After identifying the critical points, the second derivative test helps us determine whether each critical point is a relative maximum, relative minimum, or neither. This means we need the second derivative, \( f''(x) \). For \( f(x) = \sin x + \cos x \), find the second derivative by differentiating \( f'(x) = \cos x - \sin x \), which gives: \( f''(x) = -\sin x - \cos x \). Then evaluate at the critical points: if \( f''(x) \) is positive, the critical point is a relative minimum; if it's negative, the point is a relative maximum.
points of inflection
Points of inflection are where the function's concavity changes. That is, where the second derivative changes signs. For the function \( f(x) = \sin x + \cos x \), the second derivative is \( f''(x) = -\sin x - \cos x \). Set \( f''(x) \) to zero to find potential inflection points: \( -\sin x - \cos x = 0 \) which simplifies to \( \sin x = -\cos x \). Within the interval \( [0, 2\pi] \), this occurs at \( x = \frac{3\pi}{4} \) and \( x = \frac{7\pi}{4} \). These points mark where the curvature of the graph changes direction.
relative maximum
A relative maximum occurs where a function changes from increasing to decreasing. Using the second derivative test, if \( f''(x) < 0 \) at a critical point, it indicates a relative maximum. For the function \( f(x) = \sin x + \cos x \), evaluate the second derivative at the critical point \( x = \frac{\pi}{4} \): \[-\sin(\frac{\pi}{4}) - \cos(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = - \sqrt{2} < 0 \]. This confirms \( x = \frac{\pi}{4} \) is a relative maximum.
relative minimum
A relative minimum occurs where the function changes from decreasing to increasing. According to the second derivative test, if \( f''(x) > 0 \) at a critical point, that point is a relative minimum. For \( f(x) = \sin x + \cos x \), we evaluate the second derivative at the critical point \( x = \frac{5\pi}{4} \): \[-\sin(\frac{5\pi}{4}) - \cos(\frac{5\pi}{4}) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} > 0 \]. This means \( x = \frac{5\pi}{4} \) is a relative minimum.

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