Differentiation (Part-38) | S N De

Differentiation (Part-34) | S N De

 1. State Rolle's theorem. Is this theorem applicable to the function $~~f(x)=x~$ in the interval $~~-1 \leq x \leq1.$

Sol. Let $~y = f(x)~$ be a single valued function of $~x~$ defined in the closed interval $~a \leq x \leq b~$ such that

$~(i)~$ It is continuous in $~a \leq x \leq b~$

$~(ii)~$ It is differentiable in $~a < x < b~$ and

$~(iii)~~ f(a) = f(b)$

Then there exists at least one value of $~x~$ (say $~x= c~$) between $~a~$ and $~b~$ (i.e., $~a<c<b~$) such that $~f'(c) = 0~$.

2nd  Part : 

Clearly, $~f(x)=x~$ is continuous in $~ -1\leq x\leq 1~$ and differentiable in $~~-1<x<1~$. But we notice that $~~f(-1)=-1 \neq 1=f(1).$ So, the third condition of Rolle's Theorem is not satisfied. So, Rolle's Theorem is not applicable for the given function .

2. Verify the truth of Rolle's theorem for the following functions :

$~(i)~f(x)=x^2+2x-8~$ in $~~-4 \leq x \leq 2.$

Sol.  Since a polynomial function is continuous  and differentiable in any finite interval, clearly, $~f(x)~$  being a polynomial function, is continuous in $~-4 \leq x \leq 2~,$ differentiable in $~-4 < x < 2~.$ 

Also, $~~f(-4)=(-4)^2+2 \times (-4)-8=0 \\ ~~~f(2)=2^2+2 \times 2-8=0 \\ \therefore ~~f(-4)=f(2)$

So, the given function satisfies all conditions of Rolle's theorem. $~\therefore~$  there exists at least one value of $~x~$ say $~x=c \in (-4,2)~$ such that $~~f'(c)=0 \rightarrow(1)$

Now, $~~f'(x)=\frac{d}{dx}(x^2+2x-8)=2x+2 \rightarrow(2)$

So, by $~(1)~$ and $~(2)~$ we get,

$~~~~2c+2=0 \\ \therefore ~2(c+1)=0 \\ \text{or,}~~c+1=0 \\ \text{or,}~~c=-1 \in(-4,2).$

Therefore, Rolle's theorem is verified.

(ii)$~f(x)=x^3+5x^2-6x~~$ in $~~0 \leq x \leq 1.$

Sol.  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 $~0<x<1~.$

Also, $~f(0)=0^3+5\times 0^2-6\times 0=0\\~~f(1)=1^3+5\times 1^2-6 \times 1=0 \\ \therefore~~f(0)=f(1)$ 

So, the given function satisfies all conditions of Rolle's theorem. $~\therefore~$  there exists at least one value of $~x~$ say $~x=c \in (0,1)~$ such that $~~f'(c)=0 \rightarrow(1)$

Now, $~~f'(x)\\=\frac{d}{dx}(x^3+5x^2-6x)\\=3x^2+10x-6 \rightarrow(2)$

So, by $~(1)~$ and $~(2)~$ we get,

$~~3c^2+10c-6=0 \\ \therefore ~c=\frac{-10 \pm \sqrt{(10)^2-4 \times 3 \times (-6)}}{2 \times 3} \\ \text{or,}~~c=\frac{-10 \pm \sqrt{100+72}}{6} \\ \text{or,}~~c=\frac{-10 \pm \sqrt{172}}{6} \\ \text{or,}~~ c=\frac{-10 \pm 2\sqrt{43}}{6} \\ \text{or,}~~ c=\frac{-5 \pm \sqrt{43}}{3} \\ \text{or,}~~ c=\frac{-5+\sqrt{43}}{3},~\frac{-5-\sqrt{43}}{3} \\ \text{or,}~~c \approx0.52 \in (0,1) ,\\~~c \approx ~-3.85 \notin(0,1).$

Hence, there exists one value of $~c=0.52~$ in the given interval for which $~~f'(c)=0~~$ and so, Rolle's theorem is verified.

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(iii)$~f(x)=(x-1)(x-2)(x-3)~~$ in $~~[1,3].$

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

Since a polynomial function is continuous  and differentiable in any finite interval, clearly, $~f(x)~$  being a polynomial function, is continuous in $~~[1,3]~,$ differentiable in $~(1,3)~.$

Also, $~f(1)=(1-1)(1-2)(1-3)=0 \\~~f(3)=(3-1)(3-2)(3-3)=0 \\ \therefore~f(1)=f(3)$

So, the given function satisfies all conditions of Rolle's theorem. 

$~\therefore~$  there exists at least one value of $~x~$ say $~x=c \in (1,3)~$ such that $~~f'(c)=0 \rightarrow(2)$

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

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

$~~3c^2-12c+11=0 \\ \therefore~c=\frac{-(-12) \pm \sqrt{(-12)^2-4 \times 3 \times 11}}{2 \times }\\ \text{or,}~~c=\frac{12 \pm \sqrt{144-132}}{2 \times 3} \\ \text{or,}~~ c=\frac{12 \pm \sqrt{12}}{2 \times 3} \\ \text{or,}~~ c=\frac{12 \pm 2\sqrt{3}}{2\times 3} \\ \text{or,}~~ c=\frac{6+\sqrt{3}}{3} \\ \therefore~c=\frac{6+\sqrt{3}}{3} \approx 2.58 \in(1,3),\\~~c=\frac{6-\sqrt{3}}{3} \approx 1.42 \in (1,3).$

Hence, there exists  values of $~c=2.58, 1.42~$ in the given interval for which $~~f'(c)=0~~$ and so, Rolle's theorem is verified.

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(iv)$~f(x)=x^3-x^2-4x+4~~$ in $~~ -2 \leq x \leq2.$

Sol. Since a polynomial function is continuous  and differentiable in any finite interval, clearly, $~f(x)~$  being a polynomial function, is continuous in $~-2 \leq x \leq2~,$ differentiable in $~-2 < x <2~.$

Also, $~~f(-2)\\=(-2)^3-(-2)^2-4 \times (-2)+4\\=-8-4+8+4\\=0\\ ~~f(2)\\=2^3-2^2-4 \times 2+4\\=8-4-8+4\\=0 \\ \therefore~f(-2)=f(2).$

So, the given function satisfies all conditions of Rolle's theorem. 

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

Now, $~f'(x)=3x^2-2x-4\\ \therefore~f'(c)=3c^2-2c-4\\ \text{or,}~~0=3c^2-2c-4~~[\text{By (1)}] \\ \therefore c=\frac{-(-2) \pm \sqrt{(-2)^2-4\times 3 \times (-4)}}{2 \times 3} \\ \text{or,}~~ c=\frac{2 \pm \sqrt{4+48}}{6}\\ \text{or,}~~ c=\frac{2 +\sqrt{52}}{6} \approx 1.54 \in(-2,2)\\~~c=\frac{2-\sqrt{52}}{6} \approx -0.87 \in(-2,2) $

Hence, there exists  values of $~~c=1.54, -0.87~~$ in the given interval for which $~~f'(c)=0~~$ and so, Rolle's theorem is verified.

(v)$~f(x)=\sin^2x~~$ in $~~0 \leq x \leq \pi .$

Sol. Clearly $~f(x)~$ is a trigonometric function $~\sin^2x~$ is continuous in $~[0,\pi]~$ and differentiable in $~(0,\pi).$ 

Also, $~~f(0)=\sin 0 =0=\sin \pi=f(\pi).$

So, the given function satisfies all conditions of Rolle's theorem.

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

Now, $~~f'(x)=\frac{d}{dx}(\sin^2x)=2\sin x\cos x \\ \therefore f'(x)=\sin2x \\ \text{so,}~~f'(c)=\sin 2c \\ \text{or,}~~0=\sin 2c~~[\text{By (1)}]\\ \therefore \sin 2c=0=\sin \pi \\ \text{or,}~ 2c=\pi \\ \text{or,}~ c=\frac{\pi}{2} \in (0,\pi)$

Hence, there exists one value of $~c=\pi/2~$ in the given interval for which $~~f'(c)=0~~$ and so, Rolle's theorem is verified.