Examples on Graphing Using Derivatives Set 3
Example - 39
Plot the graph of \(\begin{align}f\left( x \right) = \frac{1}{2}\sin 2x + \cos x.\end{align}\)
Solution: * 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 \([0,2\pi ]\)
* \(f\left( x \right)\) is continuous and differentiable on \(\mathbb{R}\)
* There are no asymptotes to f(x)
* \(f'\left( x \right) = \cos 2x - \sin x\)
\(\begin{align} &= 1 - 2{\sin ^2}x - \sin x\\\\ &= \left( {1 + \sin x} \right)\left( {1 - 2\sin x} \right)\end{align}\)
This is 0 in \(\left[ 0,2\pi \right]\) when
\(\begin{align}x = \frac{\pi }{6},\frac{{5\pi }}{6},\frac{{3\pi }}{2}\end{align}\)
\(f\left( x \right) = - 2\sin 2x - \cos x\)
Now, \(\begin{align}\;\; f''\left( {\frac{\pi }{6}} \right)< 0,\,\,f\left( {\frac{{5\pi }}{6}} \right) > 0\end{align}\) and \(\begin{align}f\left( {\frac{{3\pi }}{2}} \right) = 0\end{align}\)
\(\begin{align} \Rightarrow \quad x = \pi /6\end{align}\) is a local maximum for \(\begin{align}f\left( x \right);f\left( {\frac{\pi }{6}} \right) = \frac{{3\sqrt 3 }}{4}\end{align}\)
\(\begin{align} x = \frac{{5\pi }}{6}\end{align}\) is a local minimum for \(\begin{align} f\left( x \right);f\left( {\frac{{5\pi }}{6}} \right) = \frac{{ - 3\sqrt 3 }}{4}\end{align}\)
\(\begin{align} x = \frac{{3\pi }}{2}\end{align}\) is a point of inflexion; \(\begin{align} f\left( {\frac{{3\pi }}{2}} \right) = 0.\end{align}\)
We now need to analyze to sign of \(f''\left( x \right).\)
\(\begin{align} &f''\left( x \right) = - 2\sin 2x - \cos x\\\\&\qquad= - 4\sin x\cos x - \cos x\\\\& \qquad= - \cos x\left( {1 + 4\sin x} \right)\end{align}\)
This is 0 in \([0,2\pi ]\) when
\[x = \frac{\pi }{2},\pi + {\sin ^{ - 1}}\frac{1}{4},\,\frac{{3\pi }}{2},2\pi - {\sin ^{ - 1}}\frac{1}{4}\]
We see that f (x) will change its convexity at four different points.
\[ \Rightarrow \quad f'' \left( x \right) > 0\,\,\forall x \in \left( {\frac{\pi }{2},\pi + {{\sin }^{ - 1}}\frac{1}{4}} \right) \cup \left( {\frac{{3\pi }}{2},2\pi - {{\sin }^{ - 1}}\frac{1}{4}} \right)\]
so that f (x) is concave upwards in these intervals
\[ \Rightarrow \quad f''\left( x \right) < 0\,\;\;\forall x \in \left( {0,\frac{\pi }{2}} \right) \cup \left( {\pi + {{\sin }^{ - 1}}\frac{1}{4},\frac{{3\pi }}{2}} \right) \cup \left( {2\pi - {{\sin }^{ - 1}}\frac{1}{4},2\pi } \right)\]
so that f (x) is concave downwards in these intervals.
\( \begin{align}* \qquad f\left( 0 \right) = 1,f\left( {\frac{\pi }{2}} \right) = 0,\,\,f\left( {2\pi } \right) = 1\end{align}\)
The graph has been plotted below for \(\left[ 0,2\pi \right]\)
Example - 40
Plot the graph of \(y = x + \ln \left( {{x^2} - 1} \right)\).
Solution:
* The Domain is given by
\(\begin{align} &\qquad\quad{x^2} - 1 > 0 \\ & \Rightarrow \quad \,D = \mathbb{R}\backslash [ - 1,\,\,1] \\ \end{align} \)
* \(f\left( x \right)\) 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
\(\begin{align}&* \qquad y'=1+\frac{2x}{{{x}^{2}}-1} \\\\ & \qquad
\quad y'=0\;\text{when}\,\,{{x}^{2}}+2x-1=0 \\\\
& \qquad\qquad\quad\quad\Rightarrow \quad x=-1\pm \sqrt{2} \\\\ &\qquad\qquad\qquad\quad \quad\;\; x=-1+\sqrt{2}\,\,\,\text{does}\,\text{not}\,\text{belong}\,\text{to}\,\text{D;} \\\\ & \qquad\quad \qquad \qquad \quad\;\; x=-1-\sqrt{2}\,\,\text{is}\,\text{an}\,\text{extremum}\,\text{point} \\\\
& \qquad \quad\quad \quad\;\; \Rightarrow \quad y''=-\frac{2\left( {{x}^{2}}+1 \right)}{{{\left( {{x}^{2}}-1 \right)}^{2}}}<0\,\,\,\,\,\forall \,x \\
\end{align}\)
\( \Rightarrow \) 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:
TRY YOURSELF - VI
Q. 1 Plot the graphs of the following functions by completely analysing their corresponding equations (as done in class; this means that you have to analyse their limits at different points as well as their first and second derivatives in different intervals):
| \((i) \begin{align}\quad y = \frac{1}{{1 - {x^2}}}\end{align}\) | \((ii)\begin{align}\quad y = \frac{x}{{{x^2} - 1}}\end{align}\) | \((iii) \begin{align}\quad y = \frac{1}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}\end{align}\) |
| \( (iv)\begin{align}\quad y = \frac{1}{x} + 4{x^2}\end{align}\) | \((v)\begin{align}\quad y = {x^2} + \frac{1}{{{x^2}}}\end{align}\) | \((vi)\begin{align}\quad y = \frac{{{x^3}}}{{3 - {x^2}}}\end{align}\) |
| \((vii)\begin{align}\quad y = \frac{{{x^3}}}{{2{{\left( {x + 1} \right)}^2}}}\end{align}\) | \((viii)\begin{align}\quad y = \frac{{{x^3}}}{{x - 1}}\end{align}\) | \( (ix)\begin{align}\quad y = \frac{{{x^4}}}{{{x^3} - 1}}\end{align}\) |
| \( (x)\begin{align}\quad y = \frac{{{{\left( {x - 1} \right)}^2}}}{{{{\left( {x + 1} \right)}^3}}}\end{align}\) | \((xi)\begin{align}\quad y = \frac{{\left( {{x^2} - 1} \right)\left( {x - 2} \right)}}{x}\end{align}\) | \( (xii)\begin{align}\quad y = \frac{x}{{{e^x}}}\end{align}\) |
| \( (xiii)\begin{align}\quad y = {x^2}{e^{ - x}}\end{align}\) | \( (xiv)\begin{align}\quad y = \frac{{{e^x}}}{x}\end{align}\) | \((xv)\quad y = x - \ln \left( {x + 1} \right)\) |
| \((xvi) \quad y = \ln \left( {{x^2} + 1} \right)\) | \((xvii)\begin{align}\quad y = {x^2}{e^{ - {x^2}}}\end{align}\) | \( (xviii) \quad y = {x^3}{e^{ - x}}\) |
| \((xix)\begin{align}\quad y = x{e^{\frac{{ - {x^2}}}{2}}}\end{align}\) | \((xx)\begin{align}\quad y = \frac{1}{{{e^x} - 1}}\end{align}\) | \((xxi)\begin{align}\quad y = x + \frac{{\ln x}}{x}\end{align}\) |
| \((xxii)\begin{align}\quad y = {\left( {1 + \frac{1}{x}} \right)^x}\end{align}\) | \( (xxiii) \begin{align}\quad y = x + \sin x\end{align}\) | \((xxiv)\quad y = \ln \cos x\) |
| \((xxv) \quad y = x\sin x\) | \( (xxvi)\quad y = x - 2{\tan ^{ - 1}}x\) | \((xxvii)\quad {y^2} = {x^3} - x\) |
| \((xxviii)\quad {y^2} = {x^3} + 1\) | \((xxix)\quad {y^2} = {x^2} - {x^4}\) | \( (xxx)\quad y = \cos x - \ln \cos x\) |
Q. 2 Find the asymptotes of the following curves and then graph them:
| \((i)\begin{align}\quad \frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\end{align}\) | \((ii)\quad xy = a\) | \((iii)\begin{align} \quad y = \frac{1}{{{x^2} - 4x + 5}}\end{align}\) |
| \( (iv)\quad {y^3} = {a^3} - {x^3}\) | \( (v)\quad y = x\,{e^x}\) | \((vi) \begin{align} \quad y = x\,{e^{\frac{2}{x}}} + 1 \end{align}\) |
| \( (vii)\quad {\left( {y + x + 1} \right)^2} = {x^2} + 1\) | \((viii)\quad x{y^2} + {x^2}y = {a^3}\) | \((ix)\quad {y^3} = 6{x^2} + {x^3}\) |
| \((x)\begin{align} \quad y = 2x + {\tan ^{ - 1}}\frac{x}{2}\end{align}\) | ||