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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:
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