# Applications of Derivatives Set 2

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Example - 3

Let   f\left( x \right) = \left\{ \begin{align}&- {x^3} + \frac{{{b^3} - {b^2} + b - 1}}{{{b^2} + 3b + 2}},\,\,\,\,0 \le x < 1\\&\qquad\qquad \quad2x - 3,\qquad\quad 1 \le x \le 3\end{align} \right\} Find all possible real values of b such that (x) has the smallest value at x = 1.

Solution: Notice that$$f\left( 1 \right) = - 1$$ (from the lower definition (x))

Also, (x) is monotonically decreasing on [0, 1) and monotonically increasing on [1, 3).

Therefore, all we require for (x) to have its minimum at x = 1 is:

$$\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) \ge f\left( 1 \right)$$ {i.e., the minimum of the left side function must not be less than (1)}

\begin{align}& \Rightarrow \qquad - 1 + \frac{{{b^3} - {b^2} + b - 1}}{{{b^2} + 3b + 2}} \ge - 1\\\\ &\Rightarrow \qquad \frac{{{b^3} - {b^2} + b - 1}}{{{b^2} + 3b + 2}} \ge 0\\\\& \Rightarrow \qquad \frac{{\left( {b - 1} \right)\left( {{b^2} + 1} \right)}}{{\left( {b + 1} \right)\left( {b + 2} \right)}} \ge 0\end{align}

Upon solving, this yields:

$b \in \left( { - 2, - 1} \right) \cup \left[ {1,\infty } \right)$

Example - 4

Using the relation $$2\left( {1 - \cos x} \right) < {x^2},x \ne 0$$ or otherwise, prove that  $$\sin \left( {\tan x} \right) \ge x$$ for all  $$x \in \left[ {0,\pi /4} \right]$$

Solution: Notice that $$'\sin \left( {\tan x} \right)'$$ and 'x' have equal values at x = 0. If we consider the function

$f\left( x \right) = \sin \left( {\tan x} \right) - x$

and try to show that it is increasing, we would obtain

\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,f\left( x \right) > f\left( 0 \right)\\&or\,\,\,\,\,\,\,\,\sin \left( {\tan x} \right) - x \ge 0\end{align}

Hence, our task could be accomplished by showing that (x) is increasing.

\begin{align}&f'\left( x \right) = \cos \left( {\tan x} \right){\sec ^2}x - 1\\\\&\,\,\,\,\,\,\,\,\,\,\,\,\; = \cos \left( {\tan x} \right)\left( {1 + {{\tan }^2}x} \right) - 1\\&\,\,\,\,\,\,\,\,\,\,\,\,\; = {\tan ^2}x\cos \left( {\tan x} \right) - \left( {1 - \cos \left( {\tan x} \right)} \right)\\\\& \qquad\; > {\tan ^2}x\cos \left( {\tan x} \right) - \frac{{{{\tan }^2}x}}{2}\;\;\;\; \left( {{\text{using the given inequality}}} \right)\\\\& \qquad\; = \frac{1}{2}{\tan ^2}x\left( {2\cos \left( {\tan x} \right) - 1} \right)\\\\ &\qquad\; = \frac{1}{2}{\tan ^2}x\left\{ {2\left( {\cos \left( {\tan x} \right) - 1} \right) + 1} \right\}\\\\&\qquad\; > \frac{1}{2}{\tan ^2}x\left( {1 - {{\tan }^2}x} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{Again using the given inequality}}} \right)\end{align}

$$For\,\,x\in \left[ 0,\pi /4 \right],\,\,\tan x\in \left[ 0,1 \right]\,so\,\,that\,\,\left( 1-{{\tan }^{2}}x \right)\ge 0$$

\Rightarrow\begin{align} \quad f'\left( x \right)>\frac{1}{2}{{\tan }^{2}}x\left( 1-{{\tan }^{2}}x \right)\ge 0\end{align}

$$\Rightarrow\quad f'\left( x \right)>0$$

$$\Rightarrow \quad f\left( x \right)\,\,is\,increasing\,on\,\,\left[ 0,\pi /4 \right]$$

$$\Rightarrow \quad \sin \left( \tan x \right)\ge \,x\,\,\forall x\in \left[ 0,\pi /4 \right]$$

Example - 5

Show that $$\cos \left( {\sin x} \right) > \sin \left( {\cos x} \right)\,\,\,\forall \,x\,\, \in \left( {0,\pi /2} \right)$$

Solution: The approach we have followed in the previous questions could be applied here to prove that $$f\left( x \right) = \cos \left( {\sin x} \right) - \sin \left( {\cos x} \right)$$is increasing. However, $$f'\left( x \right)$$, becomes complicated and proving that it is positive is not straightforward like in the previous cases (you are urged to try this out).

Instead of considering the expressions $$\cos \left( {\sin x} \right)$$ and $$\sin \left( {\cos x} \right)$$, we can consider \begin{align}\sin \left( {\frac{\pi }{2} - \sin x} \right)\,\,{\rm{and}}\,\,\,\sin \left( {\cos x} \right).\end{align} This is because ‘sin’ is a monotonically increasing function in\begin{align}\left( {0,\,\,\frac{\pi }{2}} \right)\end{align}, so that to determine the larger of the two values above, we just need to compare their arguments, i.e.\begin{align}\left( {\frac{\pi }{2} - \sin x} \right)\end{align}, and cos x

For\begin{align}\left( {0,\,\frac{\pi }{2}} \right)\end{align}

$\sin x + \cos x\,\, < \,\frac{\pi }{2}$

because the maximum value of LHS is$$\sqrt 2$$ while the RHS $$\simeq 1.57$$

\begin{align}& \Rightarrow\quad \,\,\cos x < \frac{\pi }{2}\,\,\, - \,\,\,\sin x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\,\forall \,x \in \left( {0,\frac{\pi }{2}} \right) \\\\ &\Rightarrow \quad \,\,\sin \left( {\cos x} \right) < \,\,\sin \left( {\frac{\pi }{2} - \sin x} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\forall \,x \in \left( {0,\frac{\pi }{2}} \right) \\\\ &\Rightarrow \quad \,\,\sin \left( {\cos x} \right) < \,\,\cos \left( {\sin x} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\forall \,x \in \left( {0,\frac{\pi }{2}} \right)\end{align}

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