6.12 Curve Sketching

  1. Domain: The domain is $$\begin{array}{rl}\{x|3x-6& \ne 0\}=\{x|x\ne 2\}\\ & =(-\mathrm{\infty },2)\cup (2,+\mathrm{\infty })\\ & =\mathbb{R}-\{2\}.\end{array}$$

2. Symmetry: None $$\begin{array}{rl}f(-x)& =\frac{2(-x)-1}{3(-x)-6}\\ & =\frac{-2x-1}{-3x-6}\\ & \ne f(x)\text{ or}-f(x)\end{array}$$

3. Asymptote: (a) Horizontal Asymptote. Because $$\underset{x\to \pm \mathrm{\infty }}{lim}\frac{2x-1}{3x-6}=\underset{x\to \pm \mathrm{\infty }}{lim}\frac{2x}{3x}=\frac{2}{3}$$ the line $y=2/3$ is the horizontal asymptote, and consequently, there is no oblique asymptote.

(b) Vertical Asymptote. Because $$\underset{x\to 2^{+}}{lim}\frac{2x-1}{3x-6}\stackrel{\left[\frac{3}{0^{+}}\right]}{=}+\mathrm{\infty }$$ and $$\underset{x\to 2^{-}}{lim}\frac{2x-1}{3x-6}\stackrel{\left[\frac{3}{0^{-}}\right]}{=}-\mathrm{\infty }$$ the line $x=2$ is a vertical asymptote.

4. Intercepts: (a) the $y$-intercept $$f(0)=\frac{2(0)-1}{3(0)-6}=\frac{1}{6}$$ so the graph of $f$ meets the $y$-axis at $(0,1/6)$.
(b) the $x$-intercept $$\frac{2x-1}{3x-6}=0\Rightarrow x=\frac{1}{2}.$$ Therefore, the graph of $f$ meets the $x$-axis at $(1/2,0)$.
So far we know the graph of $f$ looks like this:

 

5. Intervals of increase or decrease: $$\begin{array}{rl}{f}^{\prime }(x)& =\frac{2(3x-6)-3(2x-1)}{(3x-6{)}^2}\\ & =\frac{-9}{(3x-6{)}^2}<0.\end{array}$$ Because $f^{\prime }(x)<0$ but $f$ is discontinuous at $x=2$, we can say that $f$ is decreasing on $(-\mathrm{\infty },2)$ and on $(2,+\mathrm{\infty })$.

6. Local Maxima and Minima: There is no critical point because $f^{\prime }(x)\ne 0$. Although $f^{\prime }(2)$ does not exist, $x=2$ is not a critical point because $x=2$ is not in the domain of $f$.

7. Concavity and Inflection Points: $$\begin{array}{rl}{f}^{\prime \prime }(x)& =\frac{0-(-9)(3)(2)(3x-6)}{(3x-6{)}^4}\\ & =\frac{54}{(3x-6{)}^3}.\end{array}$$ $$f^{\prime \prime }(x)>0⟺(3x-6{)}^3>0⟺x>2,$$ and $f^{\prime \prime }(x)<0$ when $x<2$. Therefore, $f$ is concave up on $(2,+\mathrm{\infty })$ and is concave down on $(-\mathrm{\infty },2)$.
Although the sign of $f^{\prime \prime }$ changes at $x=2$, the function does not have an inflection point at $x=2$ because $x=2$ is not in the domain of $f$.

8. Sketch: We can use the information in the previous steps to finish the sketch.

1. Domain: The domain of the function is $$\begin{array}{rl}Dom(f)& =\{x|x^2-1\ge 0\}\\ & =\{x|x^2\ge 1\}\\ & =\{x|x\ge 1\text{ or }x\le -1\}\\ & =(-\mathrm{\infty },-1]\cup [1,+\mathrm{\infty }).\end{array}$$

2. Symmetry: None.

3. Asymptotes: The function does not have a vertical asymptote because it is continuous on its domain. Because $$\underset{x\to \pm \mathrm{\infty }}{lim}5x=\pm \mathrm{\infty },\underset{x\to \pm \mathrm{\infty }}{lim}3\sqrt{x^2-1}=+\mathrm{\infty }$$ we have $$\begin{array}{rl}\underset{x\to +\mathrm{\infty }}{lim}f(x)& =\underset{x\to +\mathrm{\infty }}{lim}(5x+3\sqrt{x^2-1})\\ & =+\mathrm{\infty }+\mathrm{\infty }=+\mathrm{\infty }.\end{array}$$ So the graph does not have a right-hand horizontal asymptote. To figure out if the graph has a left-hand horizontal asymptote, we need to find $\underset{x\to -\mathrm{\infty }}{lim}f(x)$, but $$\underset{x\to -\mathrm{\infty }}{lim}(5x+3\sqrt{x^2-1})=-\mathrm{\infty }+\mathrm{\infty }$$ is indeterminate. To evaluate $\underset{x\to -\mathrm{\infty }}{lim}f(x)$, we multiply $f(x)$ by its conjugate:
$$\begin{array}{rl}\underset{x\to -\mathrm{\infty }}{lim}f(x)& =\underset{x\to -\mathrm{\infty }}{lim}(5x+3\sqrt{x^2-1})\\ & =\underset{x\to -\mathrm{\infty }}{lim}((5x+3\sqrt{x^2-1})\frac{5x-3\sqrt{x^2-1}}{5x-3\sqrt{x^2-1}})\\ & =\underset{x\to -\mathrm{\infty }}{lim}\frac{25x^2-9(x^2-1)}{5x-3\sqrt{x^2-1}}\\ & =\underset{x\to -\mathrm{\infty }}{lim}\frac{16x^2(1-\frac{9}{16x^2})}{x(5-\frac{3}{x}\sqrt{x^2-1})}\\ & =\left(\underset{x\to \pm \mathrm{\infty }}{lim}16x\right)\left(\underset{x\to \pm \mathrm{\infty }}{lim}\frac{1-\frac{9}{16x}}{5+3\sqrt{\frac{x^2}{x^2}-\frac{1}{x^2}}}\right)(\sqrt{x^2}=-x\text{ for }x<0)\\ & =(-\mathrm{\infty })(\frac{1}{5+3\sqrt{1-0}})=-\mathrm{\infty }.\end{array}$$ Therefore, $f$ does not have horizontal asymptotes, but it may have oblique asymptotes. Recall that if $y=mx+b$ is an oblique asymptote then $f(x)/x\to m$ and $f(x)-mx\to b$ as $x\to +\mathrm{\infty }$ or $x\to -\mathrm{\infty }$.
Oblique asymptote as $x\to +\mathrm{\infty }$: $$\begin{array}{rl}\underset{x\to +\mathrm{\infty }}{lim}\frac{f(x)}{x}& =\underset{x\to +\mathrm{\infty }}{lim}\frac{5x+3\sqrt{x^2-1}}{x}\\ & =\underset{x\to +\mathrm{\infty }}{lim}(5+3\frac{1}{x}\sqrt{x^2-1})\\ & =\underset{x\to +\mathrm{\infty }}{lim}(5+3\sqrt{\frac{x^2}{x^2}-\frac{1}{x^2}})\\ & =5+3\sqrt{1-0}=8\end{array}$$ and $$\begin{array}{rl}\underset{x\to +\mathrm{\infty }}{lim}(f(x)-8x)& =\underset{x\to +\mathrm{\infty }}{lim}(-3x+3\sqrt{x^2-1})\\ & =\underset{x\to +\mathrm{\infty }}{lim}\left((-3x+3\sqrt{x^2-1})\frac{3x+3\sqrt{3x^2-1}}{3x+3\sqrt{x^2-1}}\right)\\ & =\underset{x\to +\mathrm{\infty }}{lim}\frac{-9x^2+3(3x^2-1)}{3x+3\sqrt{x^2-1}}\\ & =\underset{x\to +\mathrm{\infty }}{lim}\frac{-3}{3x+3\sqrt{x^2-1}}\\ & \stackrel{\left[\frac{-3}{+\mathrm{\infty }}\right]}{=}0.\end{array}$$ Therefore, $y=8x$ is an oblique asymptote as $x\to +\mathrm{\infty }$.
Oblique asymptote as $x\to -\mathrm{\infty }$ $$\begin{array}{rl}\underset{x\to -\mathrm{\infty }}{lim}\frac{f(x)}{x}& =\underset{x\to -\mathrm{\infty }}{lim}\frac{5x+3\sqrt{x^2-1}}{x}\\ & =\underset{x\to -\mathrm{\infty }}{lim}(5+3\frac{1}{x}\sqrt{x^2-1})\\ & =\underset{x\to -\mathrm{\infty }}{lim}(5-3\sqrt{\frac{x^2}{x^2}-\frac{1}{x^2}})\\ & =5-3\sqrt{1-0}=2.\end{array}$$ and $$\underset{x\to -\mathrm{\infty }}{lim}(f(x)-2x)=0.$$ Therefore, $y=2x$ is an oblique asymptote as $x\to -\mathrm{\infty }$.

4. Intercepts: The $y$-intercept: Because $x=0$ is not in the domain of $f$, the graph of $f$ does not intersect the $y$-axis.

The $x$-intercepts: $$f(x)=0$$ $$5x+3\sqrt{x^2-1}=0$$ $$5x=-3\sqrt{x^2-1}$$ Squaring each side: $$25x^2=9(x^2-1)$$ $$16x^2=-9.$$ So there is no solution and the graph of $f$ does not intersect the $x$-axis.

5. Intervals of Increase or Decrease: $$\begin{array}{rl}{f}^{\prime }(x)& =5+\frac{3(2x)}{2\sqrt{x^2-1}}\\ & =5+\frac{3x}{\sqrt{x^2-1}}.\end{array}$$ To determine the sign of $f^{\prime }$, first we find the zero(s) of $f^{\prime }$ $$f^{\prime }(x)=0$$ $$5+\frac{3x}{\sqrt{x^2-1}}=0$$ $$\frac{3x}{\sqrt{x^2-1}}=-5$$ $$\begin{array}{}\text{(*)}& 3x=-5\sqrt{x^2-1}\end{array}$$ Squaring both sides: $$9x^2=25(x^2-1)$$ $$x=\pm \sqrt{\frac{25}{16}}=\pm \frac{5}{4}.$$ We note that only $x=-5/4$ is acceptable, and $x=5/4$ is an extraneous solution (see Section 1.20), because in Equation (*), the right-hand side is negative  ($-5\sqrt{x^{2}-1}\leq0$)andsomustbetheleft-handsideand$x$.

We need to determine the sign of $f^{\prime }$ in three intervals $(-\mathrm{\infty },-5/4),(-5/4,-1)$, and $(1,+\mathrm{\infty })$. To do so, we choose a number in each interval and compute $f^{\prime }$ for that number. For example, we may choose $x=-2$ in $(-\mathrm{\infty },-5/4),$ $x=-1.2$ in the interval $(-5/4,-1)$, and $x=2$ in $(1,+\mathrm{\infty })$: $$f^{\prime }(-2)=5+\frac{3(-2)}{\sqrt{(-2{)}^2-1}}=5-\frac{6}{\sqrt{3}}\approx 1.536>0$$ $$f^{\prime }(-1.2)=5+\frac{3(-1.2)}{\sqrt{(-1.2{)}^2-1}}=5-\frac{3.6}{\sqrt{0.44}}\approx -0.427>0$$ $$f^{\prime }(2)=5+\frac{3(2)}{\sqrt{2^2-1}}=5+\frac{6}{\sqrt{3}}>0.$$ Therefore,

 

	$-\mathrm{\infty }$		$-\frac{5}{4}$		$-1$		$1$		$+\mathrm{\infty }$
sign of $f^{\prime }(x)$		$+++$	$0$	$-–-$		undef		$+++$
Increasing/Decreasing $f(x)$		$<img\; draggable="false"\; role="img"\; class="emoji"\; alt="\nearrow "\; src="https://s.w.org/images/core/emoji/15.0.3/svg/2197.svg">$	max	$<img\; draggable="false"\; role="img"\; class="emoji"\; alt="\searrow "\; src="https://s.w.org/images/core/emoji/15.0.3/svg/2198.svg">$				$<img\; draggable="false"\; role="img"\; class="emoji"\; alt="\nearrow "\; src="https://s.w.org/images/core/emoji/15.0.3/svg/2197.svg">$

6. Local Maxima and Minima: Because $x=-5/4$ is a critical number $f^{\prime }(-5/4)=0$ and the function changes from increasing to decreasing, $f(-5/4)=-4$ is a local minima.

7. Concavity and Inflection Points: $$\begin{array}{rl}{f}^{\prime \prime }(x)& =\frac{3\sqrt{x^2-1}-{\displaystyle \frac{3x^2}{\sqrt{x^2-1}}}}{x^2-1}\\ & =-\frac{3}{(x^2-1{)}^{3/2}}.\end{array}$$ Because $\sqrt{(x^2-1{)}^3}>0$, $f^{\prime \prime }(x)<0$ for every $x$ in the domain and the graph is concave down in its domain. The direction of concavity does not change, so there is no point of inflection.

8. Sketch: Using this information, we sketch the graph in the following figure.

1. Domain: The function is defined for all values of $x$ except $x=\pm 1$, which would make the denominator zero.

2. Symmetry: The function is odd: $$\begin{array}{rl}f(-x)& =\frac{-x}{\sqrt[3]{(-x{)}^2-1}}\\ & =-\frac{x}{\sqrt[3]{x^2-1}}\\ & =-f(x)\end{array}$$ and therefore the graph is symmetric about the origin. This simplifies the sketch of the graph.

3. Asymptotes: (a) Vertical asymptote: The function is discontinuous at $x=\pm 1$, so we need to find the limit of $f(x)$ as $x\to \pm 1$ to see if the function goes to infinity. Because of the symmetry of the graph, we investigate only the limit as $x\to 1$: $$\underset{x\to 1^{+}}{lim}f(x)=\underset{x\to 1^{+}}{lim}\frac{x}{\sqrt[3]{x^2-1}}\stackrel{\left[\frac{1}{0^{+}}\right]}{=}+\mathrm{\infty }$$ $$\underset{x\to 1^{-}}{lim}f(x)=\underset{x\to 1^{-}}{lim}\frac{x}{\sqrt[3]{x^2-1}}\stackrel{\left[\frac{1}{0^{-}}\right]}{=}-\mathrm{\infty }$$ (b) Horizontal asymptote: $$\begin{array}{rl}\underset{x\to +\mathrm{\infty }}{lim}\frac{x}{\sqrt[3]{x^2-1}}& =\underset{x\to +\mathrm{\infty }}{lim}\frac{x}{x^{2/3}\sqrt[3]{1-\frac{1}{x^2}}}\\ & =\underset{x\to +\mathrm{\infty }}{lim}{x}^{1/3}\underset{x\to +\mathrm{\infty }}{lim}\frac{1}{\sqrt[3]{1-\frac{1}{x^2}}}\\ & =+\mathrm{\infty }(1)=+\mathrm{\infty }.\end{array}$$ So there is no horizontal asymptote.

(c) Oblique (or Slant) Asymptote: $$m=\underset{x\to +\mathrm{\infty }}{lim}\frac{f(x)}{x}=\underset{x\to +\mathrm{\infty }}{lim}\frac{1}{\sqrt[3]{x^2-1}}\stackrel{\left[\frac{1}{+\mathrm{\infty }}\right]}{=}0.$$ and $$\underset{x\to +\mathrm{\infty }}{lim}[f(x)-mx]=\underset{x\to +\mathrm{\infty }}{lim}f(x)=+\mathrm{\infty }.$$ So there are no oblique asymptotes either.

4. Intercepts: The $x$– and $y$-intercepts are both $0$.

5. Intervals of Increase or Decrease: $$\begin{array}{rl}{f}^{\prime }(x)& =\frac{1(x^2-1{)}^{1/3}-x\left(\frac{1}{3}\right)(2x)(x^2-1{)}^{-2/3}}{(x^2-1{)}^{2/3}}\\ & =\frac{3(x^2-1)-2x^2(x^2-1{)}^{-2/3}}{3(x^2-1{)}^{2/3}}\times \frac{(x^2-1{)}^{2/3}}{(x^2-1{)}^{2/3}}\\ & =\frac{x^2-3}{3(x^2-1{)}^{4/3}}.\end{array}$$ The denominator is always nonnegative, so the sign of $f^{\prime }$ is solely determined by the sign of $(x^2-3)$. Therefore, $f^{\prime }$ is negative between $x=-\sqrt{3}$ and $x=\sqrt{3}$ and is positive everywhere else.

 

$x$	$-\mathrm{\infty }$		$-\sqrt{3}$		$\sqrt{3}$		$+\mathrm{\infty }$
sign of $f^{\prime }(x)$		$+++$	$0$	$-–-$	$0$	$+++$
Increasing/Decreasing $f(x)$		$<img\; draggable="false"\; role="img"\; class="emoji"\; alt="\nearrow "\; src="https://s.w.org/images/core/emoji/15.0.3/svg/2197.svg">$	max	$<img\; draggable="false"\; role="img"\; class="emoji"\; alt="\searrow "\; src="https://s.w.org/images/core/emoji/15.0.3/svg/2198.svg">$	min	$<img\; draggable="false"\; role="img"\; class="emoji"\; alt="\nearrow "\; src="https://s.w.org/images/core/emoji/15.0.3/svg/2197.svg">$

 

6. Local Maxima and Minima: It follows from the above table that $$f(\sqrt{3})=\frac{\sqrt{3}}{(3-1{)}^{1/3}}=\frac{3^{1/3}}{2^{1/3}}\approx 1.375$$ is a local minima, and $x=-\sqrt{3}$ gives a local maxima but we don’t need this information because we plot the graph for $x\ge 0$.

So far, we know the graph for $x\ge 0$ looks like this:

7. Concavity and Inflection Points: After some simplifications, we get $$f^{\prime \prime }(x)=\frac{2x(9-x^2)}{9(x^2-1{)}^{7/3}}$$

 

			$-3$		$-1$		$0$		$1$		$3$
sign of $x$		$--$	$-$	$--$	$-$	$--$	$0$	$++$	$+$	$++$	$+$	$++$
sign of $9-x^2$		$--$	$0$	$++$	$+$	$++$	$+$	$++$	$+$	$++$	$0$	$--$
sign of $(x^2-1{)}^{7/3}$		$++$	$+$	$++$	$0$	$--$	$-$	$--$	$0$	$++$	$+$	$++$
$\therefore$ sign of $f^{\prime \prime }$		$++$	$0$	$--$	undef	$++$	$0$	$--$	undef	$++$	$0$	$--$
concavity of $f(x)$		up	$0$	down	undef	up		down	undef	up		down

 

Because at $x=\pm 3$ the direction of concavity changes, $(3,f(3))=(3,1.5)$ and $(-3,f(-3))=(-3,-1.5)$ are points of inflection.

8. Sketch: Using this information, we sketch the curve for $x\ge 0$ and then use the symmetry about the origin as in the following figure.

First, let’s simplify:

$$\begin{array}{rl}f(x)& =\frac{{\sin }^{3}x}{1+\cos x}\\ & =\frac{\sin x{\sin }^{2}x}{1+\cos x}\\ & =\frac{\sin x(1-{\cos }^{2}x)}{1+\cos x}\\ & =\frac{\sin x(1-\cos x)(1+\cos x)}{1+\cos x}\\ & =\sin x(1-\cos x)\\ & =\sin x-\sin x\cos x\\ & =\sin x-\frac{1}{2}\sin 2x\end{array}$$ provided $1+\cos x\ne 0$. Therefore:

1. Domain: The function is defined for every $x$ such that $\cos x\ne -1$ or $$\begin{array}{rl}Dom(f)& =\{x|\cos x\ne -1\}\\ & =\{x|x\ne (2k+1)\pi ,k\in \mathbb{Z}\}\end{array}$$ 2. Symmetry: The function is odd because $$\begin{array}{rl}f(-x)& =\sin (-x)-\frac{1}{2}\sin (-2x)\\ & =-\sin x+\frac{1}{2}\sin 2x\\ & =-f(x).\end{array}$$ so we can sketch the graph of $f$ only for $x>0$ and then use the fact that its graph is symmetric about the origin.

The function is periodic with period $2\pi$ because $$\begin{array}{rl}f(x+2\pi )& =\sin (x+2\pi )+\frac{1}{2}\sin (2x+4\pi )\\ & =\sin x+\frac{1}{2}\sin 2x=f(x).\end{array}$$ So to sketch the graph of $f$, we can consider only the interval $[0,2\pi ]$ or $[-\pi ,\pi ]$. However, the interval $[-\pi ,\pi ]$ is preferred because we can use the symmetry of the graph of $f$ with respect to the origin and sketch the graph only over the interval $[0,\pi ]$.

Because $f$ has a removable discontinuity at $x=\pi$: $$\underset{x\to \pi }{lim}f(x)=\underset{x\to \pi }{lim}(\sin x-\frac{1}{2}\sin 2x)=0,$$ we can sketch the graph of $f$ for $0\le x\le \pi$ and at the end remove the point $(\pi ,0)$ from the graph.

3. Asymptotes: $f$ has no asymptotes by the virtue of continuity and periodicity.

4. Intercepts: The $y$-intercept is $f(0)=0$. To find the $x$-intecept, we set $f(x)=0$ $$f(x)=\sin x-\frac{1}{2}\sin 2x=0$$ or $$\sin x(1-\cos x)=0.$$ The solutions of $\sin x=0$ in the interval $0\le x\le \pi$ are $x=0$ and $x=\pi$, and the solution of $\cos x=1$ in the interval $0\le x<\pi$ is also $x=0$. Because $x=\pi$ is not in the domain of $f$, the curve intersects the $x$-axis only at $x=0$.

5. Intervals of Increase or Decrease $$f^{\prime }(x)=\cos x-\cos 2x.$$ To find the roots of $f^{\prime }$ $$\cos x=\cos 2x$$ or $$\begin{array}{rl}\cos x& ={\cos }^{2}x-{\sin }^{2}x\\ & =2{\cos }^{2}x-1\end{array}$$ or $$2{\cos }^{2}x-\cos x-1=0.$$ This is a quadratic equation in terms of $\cos x$. So its solutions are $$\begin{array}{rl}\cos x& =\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ & =\frac{1\pm \sqrt{1+8}}{4}\\ & =1,-\frac{1}{2}\end{array}$$ The solutions of $\cos x=1$ and $\cos x=-1/2$ in the interval $[0,\pi )$ are $$\cos x=1\iff x=0$$ $$\cos x=-\frac{1}{2}\iff x=\pi -\frac{\pi }{3}=\frac{2\pi }{3}.$$ Therefore, the roots of $f^{\prime }$ in the interval $[0,\pi )$ are $$x=0,x=\frac{2\pi }{3}.$$ The sign of $f^{\prime }$ in the intervals $[0,2\pi /3]$ and $[2\pi /3,\pi ]$ can be obtained by testing a number in each interval: $$f^{\prime }\left(\frac{\pi }{2}\right)=\cos \frac{\pi }{2}-\cos \pi =0-(-1)>0$$ and $$\begin{array}{rl}{f}^{\prime }(\pi -\frac{\pi }{6})& =\cos (\pi -\frac{\pi }{6})-\cos (2\pi -\frac{\pi }{3})\\ & =-\cos \frac{\pi }{6}-\cos \frac{\pi }{3}\\ & =-\frac{\sqrt{3}}{2}-\frac{1}{2}<0\end{array}$$

sign of $f^{\prime }(x)$	$0$	$+++$	$0$	$-–-$	$-$
Increasing/Decreasing $f(x)$		$<img\; draggable="false"\; role="img"\; class="emoji"\; alt="\nearrow "\; src="https://s.w.org/images/core/emoji/15.0.3/svg/2197.svg">$	max	$<img\; draggable="false"\; role="img"\; class="emoji"\; alt="\searrow "\; src="https://s.w.org/images/core/emoji/15.0.3/svg/2198.svg">$

 

6. Local Maxima and Minima: In the interval $[0,\pi ]$, $$\begin{array}{rl}f(2\pi /3)& =\sin \frac{2\pi }{3}-\frac{1}{2}\sin \frac{4\pi }{3}\\ & =\sin \frac{\pi }{3}-\frac{1}{2}\sin (\pi +\frac{\pi }{3})\\ & =\sin \frac{\pi }{3}-\frac{1}{2}(-\sin \frac{\pi }{3})& & (\sin (\theta +\pi )=-\sin \theta )\\ & =\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{4}=\frac{3\sqrt{3}}{4}\end{array}$$ is a local maximum.

7. Concavity and Inflection Points: $$f^{\prime }(x)=\cos x-\cos 2x$$ $$\Rightarrow f^{\prime \prime }(x)=-\sin x+2\sin 2x$$ To find the direction of concavity and inflection points, we set: $$f^{\prime \prime }(x)=-\sin x+2\sin 2x=0$$ or $$\begin{array}{rl}\sin x& =2\sin 2x\\ & =4\sin x\cos x\end{array}$$ The zeros of $f^{\prime \prime }$ are $\sin x=0$ and $\cos x=1/4$ or $$\sin x=0\iff x=0,\pi$$ $$\cos x=\frac{1}{2}\iff x=\arccos \frac{1}{4}\approx 1.31812.$$ The signs of $f^{\prime \prime }$ in the intervals$[0,\arccos \frac{1}{4}]$ and $[\arccos \frac{1}{4},\pi ]$ are obtained by testing a number in each interval: $$\begin{array}{rl}{f}^{\prime \prime }(\pi /6)& =-\sin \frac{\pi }{6}+2\sin \frac{\pi }{3}\\ & =-\frac{1}{2}+2(\frac{\sqrt{3}}{2})>0\end{array}$$ and $$\begin{array}{rl}{f}^{\prime \prime }(\pi /2)& =-\sin \frac{\pi }{2}+2\sin \pi \\ & =-1+2(0)<0\end{array}$$

 

sign of $f^{\prime \prime }(x)$	$0$	$+++$	$0$	$-–-$	$0$
Concavity of $f(x)$		Up		Down

 

8. Sketch: Using this information, we sketch the graph of $f$ for $0\le x\le \pi$ and remove the point $(\pi ,0)$ from the graph (Figure 9(a)). Then we obtain the graph of $f$ in the interval $(-\pi ,\pi )$ using the symmetry about the origin (Figure 9(b)).

(a) Graph of $f$ for $0\le x<\pi$	(b) Graph of $f$ for $-\pi <x<\pi$

Figure 9

Finally, using the periodicity of $f$, we extend the graph to obtain the complete graph of $f$ (Figure 10).