tg

tg
tgt

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
Similarly, for x < 0, we apply LMVT on [x,0] to get:
{e^x} - 1 \le \dfrac{{{e^x} - x - 1}}{x} \le 0
 \Rightarrow  \,\,\,\, {e^x} \ge x + 1\,\,;x < 0
We see that {e^x} \ge x + 1 for x \in\mathbb{R}
Post a Comment