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## Sunday, 3 August 2014

### CHAPTER 22- Worked Out Examples

 Example: 1
 Sketch the graph of $f\left( x \right) = {x^6} - 3{x^4} + 3{x^2} - 5$.
 Solution: 1
• The domain is obviously $\mathbb{R}$
• $f(x)$ is an even function
• Since $f(x)$ is a polynomial function, it is continuous and differentiable on $\mathbb{R}$.
• It is obvious that there are no asymptotes to $f(x)$
•  $f'\left( x \right) = 6{x^5} - 12{x^3} + 6x$ $= 6x\left( {{x^4} - 2{x^2} + 1} \right)$ $= 6x{\left( {{x^2} - 1} \right)^2}$ $f'\left( x \right) = 0\,\,{\rm{for}}\,\,x = 0, \pm 1$ $f''\left( x \right) = 6\,{\left( {{x^2} - 1} \right)^2} + 24{x^2}\left( {{x^2} - 1} \right)$ $= \left( {{x^2} - 1} \right)\left\{ {6\left( {{x^2} - 1} \right) + 24{x^2}} \right\}$ $= \left( {{x^2} - 1} \right)\left( {30{x^2} - 6} \right)$ $\ldots(i)$ $= 6\left( {5{x^4} - 6{x^2} + 1} \right)$ $f''\left( 0 \right) = 6,\,\,f''\left( { \pm 1} \right) = 0$
$x = 0$ is a point of local minimum and $x = \pm 1$ are points of inflexion (verify that $f''(x)$ does not change sign as $x$ crosses $\pm 1$) .
Now, $f'\left( x \right) > 0\,\,{\rm{if}}\,\,x > 0$ and $f'\left( x \right) < 0$ if $x < 0$. Therefore, $f\left( x \right)$decreases on $\left( { - \infty ,0} \right)$ and increases on $\left( {0 - \infty } \right)$ There is one more important fact we must take into account. $f''\left( x \right)$ has roots $\pm 1$ and additionally, $\pm \dfrac{1}{{\sqrt 5 }}$ (from ($i$)).
Therefore, at these four points the convexity of the graph changes:
 $\Rightarrow \,\,\,\, f''\left( x \right) > 0\,\,\,\forall \,x \in \left( { - \infty , - 1} \right) \cup \left( {\dfrac{{ - 1}}{{\sqrt 5 }},\dfrac{1}{{\sqrt 5 }}} \right) \cup (1,\infty )$so that $f(x)$ is concave upwards in these intervals $\Rightarrow \,\,\,\, f''\left( x \right) < 0\,\,\,\forall \,\,x \in \left( { - 1,\dfrac{{ - 1}}{{\sqrt 5 }}} \right) \cup \left( {\dfrac{1}{{\sqrt 5 }},1} \right)$ so that $f(x)$ is concave downwards in these intervals.
• $f\left( 0 \right) = - 5,f\left( { \pm 1} \right) = - 4,\,\,f\left( { \pm 2} \right) = 23$ Therefore one root each of $f(x)$ lies in $( - 2,- 1)$ and $(1,2)$ This information is sufficient to accurately draw the graph of the given function.
 Example: 2
 Plot the graph of $f\left( x \right) = \dfrac{1}{2}\sin 2x + \cos x$.
 Solution: 2
• The domain of $f(x)$ is $\mathbb{R}$
• $f\left( x \right)$ is periodic with period $2\pi$ and therefore we need to analyze it only in $[0,2\pi ]$
• $f(x)$ is continuous and differentiable on $\mathbb{R}$
• There are no asymptotes to $f(x)$
•  $f'\left( x \right) = \cos 2x - \sin x$ $= 1 - 2{\sin ^2}x - \sin x$ $= \left( {1 + \sin x} \right)\left( {1 - 2\sin x} \right)$ This is $0$ in $[0,2\pi ]$ when $x = \dfrac{\pi }{6},\dfrac{{5\pi }}{6},\dfrac{{3\pi }}{2}$ $f''\left( x \right) = - 2\sin 2x - \cos x$ Now, $f''\left( {\dfrac{\pi }{6}} \right) < 0,\,\,f''\left( {\dfrac{{5\pi }}{6}} \right) > 0andf''\left( {\dfrac{{3\pi }}{2}} \right) = 0$ $\Rightarrow x = \pi /6$ is a local maximum for $f\left( x \right);f\left( {\dfrac{\pi }{6}} \right) = \dfrac{{3\sqrt 3 }}{4}$ $x = \dfrac{{5\pi }}{6}$ is a local minimum for $f\left( x \right);f\left( {\dfrac{{5\pi }}{6}} \right) = \dfrac{{ - 3\sqrt 3 }}{4}$ $x = \dfrac{{3\pi }}{2}$ is a point of inflexion; $f\left( {\dfrac{{3\pi }}{2}} \right) = 0$.
We now need to analyze to sign of $f''\left( x \right)$.
 $f''\left( x \right) = - 2\sin 2x - \cos x$ $= - 4\sin x\cos x - \cos x$ $= - \cos x\left( {1 + 4\sin x} \right)$
This is $0$ in $[0,2\pi ]$ when
 $x = \dfrac{\pi }{2},\pi + {\sin ^{ - 1}}\dfrac{1}{4},\,\dfrac{{3\pi }}{2},2\pi - {\sin ^{ - 1}}\dfrac{1}{4}$
We see that $f(x)$ will change its convexity at four different points.
$\Rightarrow \,\,\,\, f''\left( x \right) > 0\,\,\forall x \in \left( {\dfrac{\pi }{2},\pi + {{\sin }^{ - 1}}\dfrac{1}{4}} \right) \cup \left( {\dfrac{{3\pi }}{2},2\pi - {{\sin }^{ - 1}}\dfrac{1}{4}} \right)$ so that $f(x)$ is concave upwards in these intervals
$\Rightarrow \,\,\,\, f''\left( x \right) < 0\,\;\;\forall x \in \left( {0,\dfrac{\pi }{2}} \right) \cup \left( {\pi + {{\sin }^{ - 1}}\dfrac{1}{4},\dfrac{{3\pi }}{2}} \right) \cup \left( {2\pi - {{\sin }^{ - 1}}\dfrac{1}{4},2\pi } \right)$.
so that $f(x)$ is concave downwards in these intervals.
• $f\left( 0 \right) = 1,f\left( {\dfrac{\pi }{2}} \right) = 0,\,\,f\left( {2\pi } \right) = 1$ The graph has been plotted below for $[0,2\pi ]$
 Example: 3
 Plot the graph of $y = x + \ln \left( {{x^2} - 1} \right)$.
 Solution: 3
• The Domain is given by
 ${x^2} - 1 > 0$ $\Rightarrow \,\,\,\, \,D = \mathbb{R}\backslash [ - 1,\,\,1]$
• $f(x)$ is continuous and differentiable on $D$
• $\mathop {\lim }\limits_{x \to {1^ + }} y = - \infty \,\,;\,\,\,\mathop {\lim }\limits_{x \to - {1^ - }} y = - \infty$ $\Rightarrow \, \,\,\, x = \pm 1$ are vertical asymptotes to the curve.
Verify that the graph has no other asymptotes
• $y' = 1 + \dfrac{{2x}}{{{x^2} - 1}}$ $y' = 0$ when ${x^2} + 2x - 1 = 0$ $\Rightarrow \,\,\,\,x = - 1 \pm \sqrt 2$ $x = - 1 + \sqrt 2$ does not belong to $D$ $x = - 1 - \sqrt 2$ is an extremum point. $\Rightarrow \,\,\,\,y'' = - \dfrac{{2\left( {{x^2} + 1} \right)}}{{{{\left( {{x^2} - 1} \right)}^2}}} < 0\,\,\,\,\,\forall \,x$
$\Rightarrow$ $\,\,\,\,$ The curve is always concave downwards so that $x = - 1 - \sqrt 2$ is a point of local maximum.
• $\mathop {\lim }\limits_{x \to + \infty } y = \infty \,\,\,;\,\,\mathop {\lim }\limits_{x \to - \infty } y = - \infty$ Based on this data, the graph can be plotted as shown below: