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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: