ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 18

Sunday, 3 August 2014

CHAPTER 18 - Worked Out Examples

Example: 1

Apply Rolle’s theorem on the following functions in the indicated intervals:

	$(a)\,\,f\left( x \right) = \sin x\,\,,x \in \left[ {0,\,\,2\pi } \right]$
	$(b)\,\,f\left( x \right) = {x^3} - x,\,\,x \in \left[ { - 1,\,\,1} \right]$

Solution: 1-(a)

We know that $f\left( x \right) = \sin x$ is everywhere continuous and differentiable. Also, $f\left( 0 \right) = f\left( {2\pi } \right) = 0$

$\Rightarrow$ From Rolle’s theorem: there exists at least one $c \in \left( {0,2\pi } \right)$ such that $f'(c) = 0$ .

In fact, from the graph we see that two such c’s exist

Solution: 1-(b)

$f\left( x \right) = {x^3} - x$ being a polynomial function is everywhere continuous and differentiable. Also,

	$f\left( { - 1} \right) = f\left( 1 \right) = 0$ .
	$\Rightarrow$ From Rolle’s theorem, these exists at least one $c$ such that $f'(c) = 0$ .

Again, we see that there are two such c’s given by $f'\left( c \right) = 0$

	$\Rightarrow \,\,\,\, 3{c^2} - 1 = 0$
	$\Rightarrow \,\,\,\, c = \pm \dfrac{1}{{\sqrt 3 }}$

Example: 2

Prove that the derivative of $f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {x\sin \dfrac{1}{x}\,\,,}\,\,\,\,{x > 0}\\ {0\,\,\,\,,}\,\,\,\,{x = 0} \end{array}} \right\}$ vanishes at an infinite number of points in $\left( {0,\dfrac{1}{\pi }} \right)$

Solution: 2

The roots of $f(x)$ are given by

	$\dfrac{1}{x} = n\pi \,\,\,;\,\,n \in\mathbb{Z}$
	$\Rightarrow \,\,\,\, x = \dfrac{1}{{n\pi }}\,\,\,;\,\,\,n \in\mathbb{Z}$	$\ldots(i)$

$f(x)$ is continuous and differentiable for all $x > 0$ .

By Rolle’s theorem, between any two successive zeroes of $f(x)$ will lie a zero $f'(x)$ . Since $f(x)$ has infinite zeroes in $\left[ {0,\dfrac{1}{\pi }} \right]$ given by ( $i$ ), $f'(x)$ will also have an infinite number of zeroes.

Example: 3

If the function $f:\left[ {0,4} \right] \to\mathbb{R}$ is differentiable, then show that ${\left( {f\left( 4 \right)} \right)^2} - {\left( {f\left( 0 \right)} \right)^2} = 8f'\left( a \right)f\left( b \right)$ for some $a,b \in \left[ {0,4} \right]$ .

Solution: 3

Applying $LMVT$ on $f (x)$ in the given interval:

There exists $a \in \left( {0,4} \right)$ such that

	$f'\left( a \right) = \dfrac{{f\left( 4 \right) - f\left( 0 \right)}}{{4 - 0}}$
	$\Rightarrow \,\,\,\, f\left( 4 \right) - f\left( 0 \right) = 4f'\left( a \right)$ for some $a \in \left( {0,4} \right)$	$\ldots(i)$

Also, since $f(x)$ is continuous and differentiable, the mean of $f(0)$ and $f(4)$ must be attained by $f(x)$ at some value of $x$ in $[0,4]$ (This obvious theorem is sometimes referred to as the intermediate value theorem).

That is, there exists $b \in [0,\,4]$ such that

	$f\left( b \right) = \dfrac{{f\left( 4 \right) + f\left( 0 \right)}}{2}$
	$\Rightarrow \,\,\,\, f\left( 4 \right) + f\left( 0 \right) = 2f\left( b \right)$ for some $b \in [0\,,4]$	$\ldots(ii)$

Multiplying ( $i$ ) and ( $ii$ ), we get the desired result.

Example: 4

Using $LMVT$ , prove that ${e^x} \ge 1 + x\,\,\,for\,\,\,x \in\mathbb{R}$

Solution: 4

Consider

	$f\left( x \right) = {e^x} - x - 1$
	$\Rightarrow \,\,\,\, f'\left( x \right) = {e^x} - 1$

Now we apply $LMVT$ on $f(x)$ for the interval $[0,x]$ , assuming $x \ge 0$ :

There exists $c \in [0,x]$ such that

	$f'\left( c \right) = \dfrac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}}$
	$= \dfrac{{\left( {{e^x} - x - 1} \right) - \left( 0 \right)}}{x}$
	$= \dfrac{{{e^x} - x - 1}}{x}$

Since $f'\left( x \right)$ is strictly increasing,

	$f'\left( 0 \right) \le f'\left( c \right) \le f'\left( x \right)$
	$\Rightarrow \,\,\,\, 0 \le \dfrac{{{e^x} - x - 1}}{x} \le {e^x} - 1$
	$\Rightarrow \,\,\,\, {e^x} \ge x + 1\,\,\,\,;x \ge 0$