3. Consider the function f (x) = sin² x + sin x on the interval (0,27).a) Find the open interval on which the function is increasing or decreasing.b) Apply the First Derivative Test to identify all local extrema.

Answered: 3. Consider the function f (x) = sin² x…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

3. Consider the function f (x) = sin² x + sin x on the interval (0,27).
a) Find the open interval on which the function is increasing or decreasing.
b) Apply the First Derivative Test to identify all local extrema.

Expert Solution

Step 1

The given function is $f (x) = \sin^{2} x + \sin x$ on the interval $(0, 2 π)$ .

a) Obtain the first derivative of the above function.

$f (x) = \sin^{2} x + \sin x \Rightarrow f' (x) = 2 \sin x \cos x + \cos x$

Identify the critical points of $f (x) = \sin^{2} x + \sin x$ in the interval $(0, 2 π)$ by solving $f' (x) = 0$ .

$f' (x) = = 0 \Rightarrow 2 \sin x \cos x + \cos x = 0 \Rightarrow \cos x (2 \sin x + 1) = 0 \Rightarrow \cos x = 0, 2 \sin x + 1 = 0 \Rightarrow x = \frac{π}{2}, x = \frac{3 π}{2}, \sin x = - \frac{1}{2} \Rightarrow x = \frac{π}{2}, x = \frac{3 π}{2}, x = \frac{7 π}{6}, x = \frac{11 π}{6}$

Hence, the function $f (x) = \sin^{2} x + \sin x$ has critical points at $x = \frac{π}{2}, x = \frac{3 π}{2}, x = \frac{7 π}{6}, x = \frac{11 π}{6}$ .

Step 2

Divide the domain $(0, 2 π)$ using the above critical points into subintervals $(0, \frac{π}{2}), (\frac{π}{2}, \frac{7 π}{6}), (\frac{7 π}{6}, \frac{3 π}{2}), (\frac{3 π}{2}, \frac{11 π}{6}), (\frac{11 π}{6}, 2 π)$ .

To identify the intervals of increasing and decreasing from the above set of intervals, evaluate the first derivative of $f (x) = \sin^{2} x + \sin x$ at an arbitrary point in each interval as follows.

For $\frac{π}{4} \in (0, \frac{π}{2})$ , we have

$\begin{array}{rcl} f' (\frac{π}{4}) & = & 2 \sin (\frac{π}{4}) \cos (\frac{π}{4}) + \cos (\frac{π}{4}) \\ \approx & 1.7 > 0 \end{array}$

Since $\begin{array}{rcl} f' (x) & > & 0 \end{array}$ for a point in $(0, \frac{π}{2})$ , $\begin{array}{rcl} f' (x) & > & 0 \end{array}$ throughout the interval $(0, \frac{π}{2})$ .

Hence, by the first derivative test, the function $f (x) = \sin^{2} x + \sin x$ is increasing on $(0, \frac{π}{2})$ .

For $π \in (\frac{π}{2}, \frac{7 π}{6})$ , we have

$\begin{array}{rcl} f' (π) & = & 2 \sin (π) \cos (π) + \cos (π) \\ = & - 1 < 0 \end{array}$

Since $\begin{array}{rcl} f' (x) & < & 0 \end{array}$ for a point in $(\frac{π}{2}, \frac{7 π}{6})$ , $\begin{array}{rcl} f' (x) & < & 0 \end{array}$ throughout the interval $(\frac{π}{2}, \frac{7 π}{6})$ .

Hence, by the first derivative test, the function $f (x) = \sin^{2} x + \sin x$ is decreasing on $(\frac{π}{2}, \frac{7 π}{6})$ .

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