# Examples on Monotonicity Set 4

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Example – 17

Prove that the function \begin{align}f\left( x \right) = \frac{x}{{\sin x}}\end{align}  is strictly increasing on \begin{align}\left( {0,\frac{\pi }{2}} \right)\end{align}.

Solution: f '(x) must be positive for the entire interval \begin{align}\left( {0,\frac{\pi }{2}} \right)\end{align} if f(x) is to be increasing on this interval.

$f'\left( x \right) = \frac{{\sin x - x\cos x}}{{{{\sin }^2}x}}$

Therefore, sin x x cos x must be positive \begin{align}\forall\;\; x \in \left( {0,\frac{\pi }{2}} \right).\end{align} Observe that it is not immediately obvious whether (sin xx cos x) will be always positive for the required interval. How do we prove this then?

Let the expression $$\left(sin\,x – x \,cos \,x \right)$$ be represented by the function g(x),

i.e.

$g\left( x \right)\text{ }=sinx\,-x\,cos\,x.$

Notice that for x = 0, g(0) = 0.

Since we have to show that \begin{align}g(x) >0\;\; \forall \;\;x \in \left( {0,\frac{\pi }{2}} \right)\end{align}, we can equivalently try to show that \begin{align} g(x) > (0)\;\;\forall x \in \left( {0,\frac{\pi }{2}} \right)\end{align} {since g(0) = 0}, which could possibly happen if g(x) is increasing on \begin{align}\left( {0,\frac{\pi }{2}} \right)\end{align}.

Hence, we analyse the sign of g'(x) in \begin{align}\left( {0,\frac{\pi }{2}} \right)\end{align}

\begin{align}&g'(x) = cos x – cos x + x sin x\\\\&\qquad= x sin x\\\\\Rightarrow \qquad &g'\left( x \right) > 0\,\,\,\,\,\forall x \in \left( {0,\pi /2} \right)\\\\\Rightarrow \qquad &g\left( x \right)\, \text{ is increasing on}\;\left( {0,\frac{\pi }{2}} \right)\\\\\Rightarrow \qquad &g\left( x \right) > g\left( 0 \right)\,\,\,\,\,\,\forall x \in \left( {0,\pi /2} \right)\\\\\Rightarrow \qquad & \sin x - x\cos x > 0\,\,\,\,\,\forall x \in \left( {0,\pi /2} \right)\\\\\Rightarrow \qquad &f'\left( x \right) > 0\;\;\;\forall x \in \left( {0,\pi /2} \right)\\\\\Rightarrow \qquad &f\left( x \right)\, \text{is strictly increasing on} \left( {0,\frac{\pi }{2}} \right)\end{align}

Example – 18

Find the bigger of the two numbers $${e^\pi }{\rm{ \,\,and \,\,}}{\pi ^e}$$.

Solution: The purpose of including this example here is to demonstrate that monotonicity can be used to determine the answers to such questions by constructing some corresponding function which  could be analysed for its interval of increase/decrease.

To determine the bigger of the two numbers $${e^\pi }{\rm{ \,\,and \,\,}}{\pi ^e}$$, we can equivalently determine the bigger of the two numbers $${e^{1/e}}\,\,and\,\,{\pi ^{1/x}}$$ (why ?). This latter alternative is helpful because we can now construct a function $${\rm{f(x) = }}{{\rm{x}}^{1/x}}$$ and analyse this for monotonicity. We can then find which of the two numbers f(e) and $${\rm{f}}(\pi )$$is larger.

\begin{align}f\left( x \right) = {x^{1/x}};x > 0\\\\ \Rightarrow \qquad \ln f\left( x \right) = \frac{1}{x}\ln x\end{align}

Differentiating both sides, we get

\begin{align}&\frac{1}{{f\left( x \right)}}.f'\left( x \right) = \frac{1}{{{x^2}}} - \frac{{\ln x}}{{{x^2}}}\\\\ \Rightarrow \qquad &f'\left( x \right) = {x^{1/x}}\frac{{\left( {1 - \ln x} \right)}}{{{x^2}}}\\\\\Rightarrow \qquad &f'\left( x \right) > 0 \qquad \quad \; \rm{if } \qquad 1– ln \,x > 0 \qquad \rm{or} \qquad x < e\\\\&\rm{and} \quad f'(x) < 0 \rm \quad{if} \qquad 1– ln \, x < 0 \qquad \rm{or} \qquad x > e \end{align}

Therefore, f (x) increases on (0, e) and decreases on  $$\left( {e,\infty } \right)$$ .

\begin{align}&\Rightarrow \qquad f\left( e \right) > f\left( \pi \right) \qquad (\text{since}\,f\,\text{decreases}\,\text{for}\,x>e\,\text{and }\pi \text{ is}\,\text{greater}\,\text{than}\,e)\\\\ &\Rightarrow \qquad {e^{1/e}} > {\pi ^{1/\pi }}\\\\&\Rightarrow \qquad {e^\pi } > {\pi ^e}\end{align}

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