Examples on Graphing Using Derivatives Set 3

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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 (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 (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 (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 (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:

 \((i)\begin{align}\quad \frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\end{align}\)  \((ii)\quad xy = a\)  \((iii)\begin{align} \quad y = \frac{1}{{{x^2} - 4x + 5}}\end{align}\)
 \( (iv)\quad {y^3} = {a^3} - {x^3}\)    \( (v)\quad y = x\,{e^x}\)    \((vi) \begin{align} \quad y = x\,{e^{\frac{2}{x}}} + 1 \end{align}\)
 \( (vii)\quad {\left( {y + x + 1} \right)^2} = {x^2} + 1\)  \((viii)\quad x{y^2} + {x^2}y = {a^3}\)  \((ix)\quad {y^3} = 6{x^2} + {x^3}\)
\((x)\begin{align} \quad y = 2x + {\tan ^{ - 1}}\frac{x}{2}\end{align}\)

 

 

 

Download practice questions along with solutions for FREE:
Applications of Derivatives
grade 11 | Questions Set 1
Applications of Derivatives
grade 11 | Answers Set 1
Applications of Derivatives
grade 11 | Questions Set 2
Applications of Derivatives
grade 11 | Answers Set 2