Differentiation (Part-38) | S N De

Differentiation (Part-38) | S N De

 

$~8.~$ Verify the truth of Lagrange's mean value theorem formula for the following functions :

$~(i)~~f(x)=x^2-4x-3~$ in the interval $~~0 \leq x \leq 4.$

Solution. 

Since a polynomial function is continuous  and differentiable in any finite interval, clearly, $~f(x)~$  being a polynomial function, is continuous in $~~[0,4]~,$ differentiable in $~(0,4)~.$

Now, $~f(x)=x^2-4x-3 \\ \therefore~f'(x)=2x-4 \rightarrow(1)$

So, the given function satisfies both the conditions of Lagrange's mean value theorem. 

$~\therefore~$  there exists at least one value of $~x~$ say $~x=c \in (0,4)~$ such that 

$~~f'(c)=\frac{f(4)-f(0)}{4-0} \rightarrow(2)$

Now, $~f(4)=4^2-4 \times 4-3=-3,\\~f(0)=0^2-4 \times 0-3=-3\rightarrow(3)$

Hence by $~(1),~(2),~(3)~$ we get,

$~~2c-4=\frac{0-0}{4}\\ \text{or,}~~2c-4=0 \\ \text{or,}~~2c=4 \\ \therefore~~c=\frac 42=2.$

Since $~~c =2\in (0,4),~~$ Lagrange's mean value theorem is verified.

$~(ii)~~f(x)=x^2+x-1~~$ in $~~[0,4]$

Solution. 

Since a polynomial function is continuous  and differentiable in any finite interval, clearly, $~f(x)~$  being a polynomial function, is continuous in $~~[0,4]~,$ differentiable in $~(0,4)~.$

Now, $~f(x)=x^2+x-1 \\ \therefore~f'(x)=2x+1 \rightarrow(1)$

So, the given function satisfies both the conditions of Lagrange's mean value theorem. 

$~\therefore~$  there exists at least one value of $~x~$ say $~x=c \in (0,4)~$ such that 

$~~f'(c)=\frac{f(4)-f(0)}{4-0} \rightarrow(2)$

Now, $~f(4)=4^2+4-1=19,\\~f(0)=0^2+0-1=-1\rightarrow(3)$

Hence by $~(1),~(2),~(3)~$ we get,

$~~2c-4=\frac{19-(-1)}{4}\\ \text{or,}~~2c+1=\frac{20}{4} \\ \text{or,}~~2c+1=5  \\ \text{or,}~~ 2c=5-1\\ \therefore~~ c=\frac 42=2.$

Since $~~c =2\in (0,4),~~$ Lagrange's mean value theorem is verified.

$~(iii)~~f(x)=\sqrt{x^2-4}~~$ in $~~[2,4].$

Solution. 

Let $~f(x)~$ be a function defined on $~[a,b]~$ such that 

$~(i)~~f(x)~$ is continuous on $~[a,b],$

$~(ii)~~f(x)~$ is differentiable on $~(a,b).$

$~\therefore~$  there exists at least one value of $~x~$ say $~x=c \in (a, b)~$ such that

$~~f'(c)=\frac{f(b)-f(a)}{b-a} \rightarrow(1)$

Now, $~~f(x)=\sqrt{x^2-4}~~~\text{(given)},\\~~f'(x)\\=\frac{d}{dx}(\sqrt{x^2-4})\\=\frac{d}{dx}~[(x^2-4)^{1/2}]\\=\frac 12(x^2-4)^{\frac 12-1}~\cdot \frac{d}{dx}(x^2-4)\\=\frac{1}{2\sqrt{x^2-4}}~\cdot 2x\\=\frac{x}{\sqrt{x^2-4}}\rightarrow(2)$

The function $~f(x)~$ satisfies the two conditions of Lagrange's mean value theorem.

So, to find the value of $~c~$, we can use the formula as mentioned in $~(1),~$ 

and get $~~f'(c)\\=\frac{f(4)-f(2)}{4-2}\\=\frac 12 [\sqrt{4^2-4}-\sqrt{2^2-4}]\\ =\frac{\sqrt{12}}{2} \\ \therefore~\frac{c}{\sqrt{c^2-4}}=\frac{\sqrt{12}}{2} \\ \text{or,}~~\frac{c^2}{c^2-4}=\left(\frac{\sqrt{12}}{2}\right)^2\\ \text{or,}~~\frac{c^2}{c^2-4}=\frac{12}{4}=3 \\ \text{or,}~~ 3(c^2-4)=c^2 \\ \text{or,}~~ 3c^2-c^2=12 \\ \text{or,}~~ 2c^2=12 \\ \text{or,}~~ c^2=\frac{12}{2}=6 \\ \text{or,}~~ c=\sqrt{6} \approx2.45  $

$~\therefore~$  there exists at least one value of  $~c=2.45\in (2,4)~$ such that $~~f'(2.45)=\frac{f(4)-f(2)}{4-2},~~$ hence Lagrange's mean value theorem is verified. 

Mathematics by S.N.DE for class 12  (WBHS,WBJEE, JEE- Main, JEE- Advanced) Paperback

$~(iv)~~2\sin x+\sin2x~~$ in $~0 \leq x \leq \pi.$

Solution. 

Let $~f(x)~$ be a function defined on $~[a,b]~$ such that 

$~(i)~~f(x)~$ is continuous on $~[a,b],$

$~(ii)~~f(x)~$ is differentiable on $~(a,b).$

$~\therefore~$  there exists at least one value of $~x~$ say $~x=c \in (a, b)~$ such that

$~~f'(c)=\frac{f(b)-f(a)}{b-a} \rightarrow(1)$

let $~~~f(x)=2\sin x+\sin2x~~$ in $~~0 \leq x \leq \pi.$

Now, $~f(x)~$ is the addition of two trigonometric functions , $~f(x)~$ is continuous on $~[0,\pi].$

Again, $~~f'(x)=\frac{d}{dx}(2\sin x+\sin 2x) \\ \text{or,}~~f'(x)=2\cos x+2\cos2x \rightarrow(2)$

Clearly, $~f'(x)~$ exists everywhere in $~(0,\pi).$

The function $~f(x)~$ satisfies the two conditions of Lagrange's mean value theorem.

So, to find the value of $~c~$, we can use the formula as mentioned in $~(1),~$ 

$~\therefore~$  there exists at least one value of $~x~$ say $~x=c \in (0,\pi)~$ such that $~~f'(c)=\frac{f(\pi)-f(0)}{\pi-0} \rightarrow(2)$

Also, $~f(\pi)=2\sin \pi+\sin2\pi=0,\\~~f(0)=2\sin 0+\sin(2 \times 0)=0 \rightarrow(3)$

So, from $~(1),~(2)~,(3)~$ we get,

$~~f'(c)=\frac{0-0}{\pi-0} \\ \text{or,}~~2\cos c+2\cos 2c=0 \\ \text{or,}~~ 2\cos c+2\cos^2c-1=0 \\ \text{or,}~~ 2\cos^2c+2\cos c-1=0 \\ \text{or,}~~ 2\cos^2c+2\cos c-\cos c-1=0 \\ \text{or,}~~ 2\cos c(\cos c+1)-(\cos c+1)=0 \\ \text{or,}~~(\cos c+1)(2\cos c-1)=0 \\ \therefore \cos c+1=0 \\~~ \Rightarrow \cos c=-1=\cos \pi \\ \therefore c =\pi   \\ ~~\text{Again,}~~2\cos c-1=0 \\ \text{or,}~~\cos c =\frac 12=\cos(\pi/3) \\ \therefore~ c=\frac{\pi}{3} \in(0, \pi) $

Hence, there exists a value of $~c=\frac{\pi}{3}$ in the given interval for which $~~f'(\pi/3)=\frac{f(\pi)-f(0)}{\pi-0}~~$ and so, Lagrange's mean value theorem is verified.

$~(v)~~f(x)=(x-1)(x-2)(x-3)~~$ in $~~0 \leq x  \leq 4.$

Solution. 

 $~~f(x)\\=(x-1)(x-2)(x-3)\\=(x^2-2x-x+2)(x-3)\\=(x^2-3x+2)(x-3)\\=x^3-3x^2+2x-3x^2+9x-6\\=x^3-6x^2+11x-6 \rightarrow(1)$

Since a polynomial function is continuous  and differentiable in any finite interval, clearly, $~f(x)~$  being a polynomial function, is continuous in $~~[0,4]~,$ differentiable in $~(0,4)~.$

Also, $~f(0)=(0-1)(0-2)(0-3)=-6 ,\\~~f(4)=(4-1)(4-2)(4-3)=6 \rightarrow(2)$

So, the given function satisfies all conditions of Lagrange's mean value theorem. 

$~\therefore~$  there exists at least one value of $~x~$ say $~x=c \in (0,4)~$ such that $~~f'(c)=\frac{f(4)-f(0)}{4-0} \rightarrow(3)$

Now, $~~f'(x)\\=\frac{d}{dx}(x^3-6x^2+11x-6)\\=3x^2-12x+11 \rightarrow(4)$

So, by $~(2),~(3)~$ and $~(4)~$ we get,

$~~3c^2-12c+11=\frac{6-(-6)}{4-0} \\ \text{or,}~~3c^2-12c+11=\frac{12}{4}=3 \\ \text{or,}~~ 3c^2-12c+11-3=0 \\ \text{or,}~~ 3c^2-12c+8=0 \\ \therefore c=\frac{-(-12) \pm \sqrt{(-12)^2 -4 \times 3 \times 8}}{2 \times 3} \\ \text{or,}~~ c=\frac{12 \pm \sqrt{144-96}}{6} \\ \text{or,}~~ c=\frac{12 \pm \sqrt{48}}{6} \\ \text{or,}~~ c=\frac{12 \pm 4 \sqrt{3}}{6} \\ \text{or,}~~ c=2 \pm \frac{2}{\sqrt{3}} \in (0,4)\rightarrow(5)$

Hence, by $~(5)~$, we can conclude  Lagrange's mean value theorem  is verified.

$~(vi)~~f(x)=x^3-2x^2-x+3~~$ in $~~[0,1].$

Solution. 

Since a polynomial function is continuous  and differentiable in any finite interval, clearly, $~f(x)~$  being a polynomial function, is continuous in $~~0 \leq x \leq 1~,$ differentiable in $~0<x<1~.$

So, the given function satisfies all conditions of Lagrange's mean value theorem. 

$~\therefore~$  there exists at least one value of $~x~$ say $~x=c \in (0,1)~$ such that $~~f'(c)=\frac{f(1)-f(0)}{1-0} \rightarrow(1)$

Now, $~~f(x)=x^3-2x^2-x+3 \\ \therefore f'(x)=3x^2-4x-1 \\ \text{so,}~~f'(c)=3c^2-4c-1 \rightarrow(2)$

Hence, by $~(1)~,(2)~$ we get,

$~~3c^2-4c-1\\=f(1)-f(0)\\=(1^3-2 \times 1^2-1+3)-(0-0-0+3)\\=(1-2-1+3)-3\\=-2 \\ \therefore~~3c^2-4c-1+2=0 \\ \text{or,}~~ 3c^2-4c+1=0 \\ \text{or,}~~ 3c^2-3c-c+1=0 \\ \text{or,}~~ 3c(c-1)-1(c-1)=0 \\ \text{or,}~~ (c-1)(3c-1)=0 \\ \text{or,}~~c-1=0,~~3c-1=0 \\ \therefore ~c=1,\frac 13\rightarrow(3)$

Hence, by $~(3)~$,we notice that there exists a value of $~c=\frac 13$ in the given interval for which $~~f'(\frac 13)=\frac{f(1)-f(0)}{1-0}~~$ and so, Lagrange's mean value theorem is verified.

$~(vii)~~f(x)=x(x-1)(x-2)~~$ in $~~\left[0,\frac 12\right]$

Solution. 

$~~f(x)\\=x(x-1)(x-2)\\=x(x^2-2x-x+2)\\=x(x^2-3x+2)\\=x^3-3x^2+2x \rightarrow(1)$

Since a polynomial function is continuous  and differentiable in any finite interval, clearly, $~f(x)~$  being a polynomial function, is continuous in $~~0 \leq x \leq \frac 12~,$ differentiable in $~0<x<\frac 12~.$

So, the given function satisfies all conditions of Lagrange's mean value theorem. 

$~\therefore~$  there exists at least one value of $~x~$ say $~x=c \in (0,1/2)~$ such that 

$~~f'(c)=\frac{f(1/2)-f(0)}{1/2-0} \rightarrow(2)$

Now, $~~f'(x)=\frac{d}{dx}(x^3-3x^2+2x)~~[\text{By (1)}] \\ \text{or,}~~f'(x)=3x^2-6x+2 \\ \therefore~~f'(c)=3c^2-6c+2\rightarrow(3)$

Hence, by $~(2)~,(3)~$ we get,

$~~3c^2-6c+2\\=2[f(1/2)-f(0)]\\=2~\times \left[\frac 12(\frac 12-1)(\frac 12-2)-0\right]\\=(-\frac 12) \times (-\frac 32) \\ \therefore~3c^2-6c+2=\frac 34 \\ \text{or,}~~4(3c^2-6c+2)=3 \\ \text{or,}~~ 12c^2-24c+8-3=0 \\ \text{or,}~~ 12c^2-24c+5=0 \\ \therefore~~ c\\=\frac{-(-24) \pm \sqrt{(-24)^2-4 \times 12 \times 5}}{2 \times 12}\\=\frac{24 \pm \sqrt{24^2- 48 \times 5}}{24}\\=1 \pm \frac{\sqrt{24(24-10)}}{24}\\=1 \pm \frac{\sqrt{24 \times 14}}{24}\\=1 \pm \frac{6\sqrt{21}}{24}\\=1 \pm \frac{\sqrt{21}}{6}\\ \approx 1 \pm 0.76 \rightarrow(4)$

Hence, by $~(4)~$,we notice that there exists a value of $~c=1-\frac{\sqrt{21}}{6} \in (0,1/2)~~$ in the given interval for which $~~f' \left(1-\frac{\sqrt{21}}{6}\right)=\frac{f(1/2)-f(0)}{\frac 12-0}~~$ and so, Lagrange's mean value theorem is verified.

$~(viii)~~f(x)=x(x+4)^2~~$ in $~~ 0\leq x \leq 4$

Solution. 

$~~f(x)\\=x(x+4)^2\\=x(x^2+2 \times x \times 4+4^2)\\=x(x^2+8x+16)\\=x^3+8x^2+16x \rightarrow(1)$

Since a polynomial function is continuous  and differentiable in any finite interval, clearly, $~f(x)~$  being a polynomial function, is continuous in $~~0 \leq x \leq 4~,$ differentiable in $~0<x<4~.$

So, the given function satisfies all conditions of Lagrange's mean value theorem. 

$~\therefore~$  there exists at least one value of $~x~$ say $~x=c \in (0,4)~$ such that 

$~~f'(c)=\frac{f(4)-f(0)}{4-0} \rightarrow(2)$

Now, $~~f'(x)=\frac{d}{dx}(x^3+8x^2+16x)~~[\text{By (1)}] \\ \text{or,}~~f'(x)=3x^2+16x+16 \\ \therefore~~f'(c)=3c^2+16c+16\rightarrow(3)$

$~~f(4)=4^3+8 \times 4^2+16 \times 4=256, \\~~f(0)=0 \rightarrow(4)$

Hence, by $~(2),(3),(4)~$ we get,

$~~3c^2+16c+16\\=\frac 14~[f(4)-f(0)]\\=\frac 14 \times 256 \\ \therefore 3c^2+16c+16=64 \\ \text{or,}~~3c^2+16c+16-64=0 \\ \text{or,}~~ 3c^2+16c-48=0 \\ \therefore~~ c\\=\frac{-16 \pm \sqrt{16^2-4 \times 3 \times (-48)}}{2 \times 3}\\=\frac{-16 \pm \sqrt{256+576}}{6}\\=\frac{-16 \pm \sqrt{832}}{6}\\=\frac{-16 \pm 8\sqrt{13}}{6}\\=-\frac{16}{6} \pm \frac{8\sqrt{13}}{6}\\=-\frac 83+\frac 43 \sqrt{13},~-\frac 83 -\frac 43 \sqrt{13}\rightarrow(5)$

Hence, by $~(5)~$,we notice that there exists a value of $~c=-\frac 83+\frac 43 \sqrt{13} \in (0,4)~~$ in the given interval for which $~~f' \left(-\frac 83+\frac 43 \sqrt{13}\right)=\frac{f(4)-f(0)}{4-0}~~$ and so, Lagrange's mean value theorem is verified.

$~9.~$ If $~~f'(x)=0~~$ for all values of $~x~$ in an interval, then show that $~f(x)~$ is constant in that interval.

Solution. 

Given that $~f'(x)=0~~\forall \in I\\ \text{or,}~~ \frac{d}{dx}(f(x))=0=\frac{d}{dx}(c)\\~~~~~~~~~~~~~~~~~~~~c \rightarrow \text{constant} \\ \therefore~ f(x)=c$

Note : $~~\forall \rightarrow\text{for all},~I \rightarrow \text{interval}$

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