Differentiation (Part-38) | S N De

Differentiation (Part-37) | S N De

 

$~4.~$ Lagrange's mean value theorem is ,

$~~f(b)-f(a)=(b-a)f'(c),~~a<c<b$

$~(i)~$ if $~f(x)=\sqrt{x}~$ and $~a=4,~b=9,~$ find $~c.$

Solution. 

$~~f(x)=\sqrt{x}=x^{\frac 12} \\ \therefore~f'(x)=\frac{d}{dx}(x^{\frac 12})=\frac 12x^{\frac 12-1}=\frac{1}{2\sqrt{x}}\rightarrow(1)$

Now, by Lagrange's mean value theorem ,

$~~f(b)-f(a)=(b-a)f'(c)~~[~c \in(a,b)] \\ \therefore~f(9)-f(4)=(9-4)f'(c)\\ \text{or,}~~\sqrt{9}-\sqrt{4}=5~\cdot \frac{1}{2\sqrt{c}}~~[\text{By (1)}] \\ \text{or,}~~ 3-2=\frac{5}{2\sqrt{c}} \\ \text{or,}~~ 2\sqrt{c}=5 \\ \text{or,}~~ (2\sqrt{c})^2=5^2 \\ \text{or,}~~ 4c=25 \\ \therefore~c=\frac{25}{4}=6.25~~\text{(ans.)}$

$~4.~$ Lagrange's mean value theorem is ,

$~~f(b)-f(a)=(b-a)f'(c),~~a<c<b$

$~(ii)~$ if $~f(x)=Ax^2+Bx+C~$ in $~a \leq x \leq b,~$ find $~c.$

Solution.

 $~~f(x)=Ax^2+Bx+C \\ \therefore~f'(x)\\=\frac{d}{dx}(Ax^2+Bx+C)\\=2Ax+B\rightarrow(1)$

Now, by Lagrange's mean value theorem ,

$~~f(b)-f(a)=(b-a)f'(c)~~[~c \in(a,b)] \\ \therefore~(Ab^2+Bb+C)-(Aa^2+Ba+C)\\~~~~=(b-a)f'(c) \\ \text{or,}~~[A(b^2-a^2)+B(b-a)]\\~~~~=(b-a)~\cdot (2Ac+B)~~[\text{By (1)}] \\ \text{or,}~~ A(b+a)(b-a)+B(b-a)\\~~~~=(2Ac+B)(b-a) \\ \text{or,}~~ 2Ac+B=\frac{(b-a)[A(b+a)+B]}{b-a} \\ \text{or,}~~ 2Ac+B=A(b+a)+B \\ \text{or,}~~ 2Ac=A(b+a) \\ \text{or,}~~ 2c=b+a \\ \text{or,}~~ c=\frac{a+b}{2}~~\text{(ans.)}$

$~5.~$ Is Lagrange's mean value theorem applicable to the function $~~f(x)=4(6-x)^{\frac 23}~$ in the interval $~~5 \leq x \leq 7~?$

Solution. 

Lagrange's mean value theorem states that if $~f(x)~$ be continuous on $\,[a, b]\,$ and differentiable on $\,(a, b)\,$ then there exists some $~c \in (a, b)~$ such that $~f'(c)=\frac{f(b)-f(a)}{b-a} \rightarrow(1)$

Now, $~~f(x)=4(6-x)^{\frac 23} \\ \therefore~f'(x)\\=4 \cdot \frac 23 \cdot (6-x)^{\frac 23-1}~\cdot \frac{d}{dx}(6-x)\\=\frac 83~\cdot(6-x)^{-\frac 13} \cdot (-1)\\=-\frac{8}{3(6-x)^{\frac 13}} \rightarrow(2)$

From $~(2),~$ we notice that $~f'(x)~$ does not exist at $~x=6 \in (5,7).$

Therefore, Lagrange's mean value theorem $~[(1)]~$is not applicable for $~f(x)~$ in the given interval.

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

$~6.~$ If $~f'(x)~$ exists and if $~~f'(x)<0~$ everywhere in the interval $~~a \leq x \leq b,~$ then show that $~f(x)~$ is a decreasing function in $~~a \leq x<b.$

Solution. 

Lagrange's mean value theorem states that if $~f(x)~$ be continuous on $\,[a, b]\,$ and differentiable on $\,(a, b)\,$ then there exists some $~c \in (a, b)~$ such that $~f'(c)=\frac{f(b)-f(a)}{b-a} \rightarrow(1)$

Now, $~~f'(c) <0~~~~[~~\text{where}~~ c \in [a,b]~~] \\ \text{or,}~~\frac{f(b)-f(a)}{b-a}<0 \rightarrow(2)\\~~~~~~~~~~~~[\text{By (1)}]$

Now, since $~~a<b,~~(b-a)>0 \rightarrow(3)$

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

$~~f(b)-f(a)<0 \\ \Rightarrow f(b)<f(a) \rightarrow(4)$

Hence, from $~(4)~$ it follows that $~f(x)~$ is a decreasing function in $~~a \leq x<b.$

$~7.~$ Form Lagrange's mean value theorem formula for the function $~~f(x)=\sin x~$ in the interval $~~x_1 \leq x \leq x_2.$

Solution. 

Lagrange's mean value theorem states that if $~f(x)~$ be continuous on $\,[a, b]\,$ and differentiable on $\,(a, b)\,$ then there exists some $~c \in (a, b)~$ such that $~f'(c)=\frac{f(b)-f(a)}{b-a} \rightarrow(1)$

Now, $~f(x)~$ being a trigonometric function is continuous in $~[x_1,x_2]~$ and differentiable in $~(x_1,x_2)~$. 

$~~f(x)=\sin x \\  \Rightarrow f'(x)=\frac{d}{dx}(\sin x)=\cos x\rightarrow(2)$

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

$~~~\cos c=\frac{f(x_2)-f(x_1)}{x_2-x_1} \\ \therefore~f(x_2)-f(x_1)=(x_2-x_1)f'(c)~~\\ ~~\text{where}~~x_1<c<x_2.$

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