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Sunday, 3 August 2014

CHAPTER 12- Worked Out Examples

   Example: 1    

Determine the intervals in which the following functions are increasing or decreasing:
(a) f\left( x \right) = {x^3} - x
(b) f\left( x \right) = {x^3} - 6{x^2} + 11x - 6
Solution: 1

In this and subsequent questions where we are required to find out the intervals of increase/decrease, we first determine f ‘(x). f(x) increases in all intervals where f'(x) > 0 and decreases in all intervals where f'(x) < 0.
(a) \,\, f'\left( x \right) = 3{x^2} - 1
Interval(s) of strict increase: \,\,\, f'\left( x \right) > 0
 \Rightarrow \,\,\,3{x^2} - 1 > 0
 \Rightarrow \,\,\,x < \dfrac{{ - 1}}{{\sqrt 3 }}{\rm{ or }}x > \dfrac{1}{{\sqrt 3 }}
Interval(s) of strict decrease: \,\,\, f'\left( x \right) < 0
 \Rightarrow \,\,\,3{x^2} - 1 < 0
 \Rightarrow \,\,\,\dfrac{{ - 1}}{{\sqrt 3 }} < x < \dfrac{1}{{\sqrt 3 }}
Therefore, f(x) increases in \left( { - \infty ,\dfrac{{ - 1}}{{\sqrt 3 }}} \right) \cup \left( {\dfrac{1}{{\sqrt 3 }},\infty } \right) and decreases in \left( {\dfrac{{ - 1}}{{\sqrt 3 }},\dfrac{1}{{\sqrt 3 }}} \right). The graph for f(x)confirms this: (to plot the graph, the knowledge of roots of f(x) helps, which is easy to obtain for this example; {x^3} - x = 0 \Rightarrow x = 0, \pm 1)
Solution: 1

f'\left( x \right) = 3{x^2} - 12x + 11
The roots of f ‘(x) are \dfrac{{12 \pm \sqrt {12} }}{6} = \dfrac{{6 \pm \sqrt 3 }}{3} = 2 \pm \dfrac{1}{{\sqrt 3 }}
Interval(s) of strict increase: \,\,\, f'\left( x \right) > 0
 \Rightarrow \,\,\,3{x^2} - 12x + 11 > 0
 \Rightarrow \,\,\,x < 2 - \dfrac{1}{{\sqrt 3 }}{\rm{  or }}x > 2 + \dfrac{1}{{\sqrt 3 }}
Interval(s) of strict decrease: \,\,\, f'\left( x \right) < 0
 \Rightarrow \,\,\,3{x^2} - 12x + 11 < 0
 \Rightarrow \,\,\,2 - \dfrac{1}{{\sqrt 3 }} < x < 2 + \dfrac{1}{{\sqrt 3 }}
f(x) can be factorised as (x - 1)(x - 2)(x - 3) so that the roots of f(x)are x = 1,2,3. The graph for f(x) is approximately sketched below:
     Example: 2     

Determine the values of x for which f(x) = {x^x},x > 0 is increasing or decreasing.
Solution: 2

To find f'(x), we first take the logarithm of both sides of the given equation:
\ln \left( {f\left( x \right)} \right) = x\ln x
Differentiating both sides, we get:
\dfrac{1}{{f\left( x \right)}} \cdot f'\left( x \right) = x \cdot \dfrac{1}{x} + \ln x \cdot 1
 = 1 + \ln x
 \Rightarrow  \,\,\,\, f'\left( x \right) = {x^x}\left( {1 + \ln x} \right)
Interval(s) of strict increase: \,\,\, f'\left( x \right) > 0
 \Rightarrow  \,\,\,\, {x^x}\left( {1 + \ln x} \right) > 0
 \Rightarrow  \,\,\,\, 1 + \ln x > 0
 \Rightarrow  \,\,\,\, x > {e^{ - 1}} = \dfrac{1}{e}
Interval(s) of strict decrease: \,\,\, f'\left( x \right) < 0
 \Rightarrow  \,\,\,\, {x^x}\left( {1 + \ln x} \right) < 0
 \Rightarrow  \,\,\,\, x < \dfrac{1}{e}
To plot the graph of f(x), notice that \mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} {x^x}
 = \mathop {\lim }\limits_{x \to 0} {e^{x\ln x}} = {e^{\mathop {\lim }\limits_{x \to 0} .x\ln x}} = {e^0} = 1.
Also,
\mathop {\lim }\limits_{x \to \infty } \left( {f\left( x \right)} \right) = \infty .
f(x) decreases in {\rm{(0, 1/e) }} and increases in \left( {\dfrac{1}{e},\infty } \right).
     Example: 3    

Separate the interval [0,\pi /2] into sub-intervals in which f\left( x \right) = {\sin ^4}x + {\cos ^4}x is increasing or decreasing.
Solution: 3

f'\left( x \right) = 4{\sin ^3}x\cos x - 4{\cos ^3}x\sin x
 = 4\sin x\cos x\left( {{{\sin }^2}x - {{\cos }^2}x} \right)
 = 2\sin 2x\left( { - \cos 2x} \right)
 =  - \sin 4x
We now need to consider the sign of f'(x) in the interval [0, p/2].
Interval (s) of strict increase: \,\,\, f'\left( x \right) > 0
 \Rightarrow  \,\,\,\,  - \sin 4x > 0
 \Rightarrow  \,\,\, \,\sin 4x < 0
 \Rightarrow  \,\,\,\, \pi  < 4x < 2\pi \,\,\,\left( \begin{array}{l}  {\rm{This\, range\, of\, 4}}x\,{\rm{ will\, ensure \,}}\\  {\rm{that\, }}x\;{\rm{itself \,lies\, in\, }}\left[ {0,\dfrac{\pi }{2}} \right] \end{array} \right)
 \Rightarrow  \,\,\,\, \dfrac{\pi }{4} < x < \dfrac{\pi }{2}
Interval(s) strict decrease:
f'\left( x \right) < 0
 \Rightarrow  \,\,\,\,  - \sin 4x < 0
 \Rightarrow  \,\,\,\, 0 < 4x < \pi
 \Rightarrow  \,\,\,\, 0 < x < \pi /4
Therefore, f(x) decreases in [0,\pi /4]and increases in . The minimum value in [0,\pi /2] is at x = \dfrac{\pi }{4} equal to f\left( x \right) = \dfrac{1}{2} and the maximum value is at x = 0 or x = \dfrac{\pi }{2} equal to f(x) = 1. The graph is approximately sketched below:
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