Question

Verify rolle's theorem

f(x)=(x-a)^m(x-b)^n,where m,n are positive intigers, in the interval [a,b]

Answer 1

Rolle's theorem state that any real value differentiable function which gives equal value at two distinct points will have a point in between them where $f^{'}(x) =0$ exist .

Now given function is $f(x)=(x-a)^{m}(x-b)^{n}$ , where m , n are positive integers , in the interval [a,b].

Now putting $x=a,$ we get $f(a)=0$

putting $x=b$ , we get $f(b)=0$ , so hence it proves that rolle's theorem satisfy .

Now , we have to check $f^{'}(x)$ is 0 between the point a and b respectively .

Now , in this case we will two functions the first one will be $(x-a)^{m}$ and the second one will be $(x-b)^{n}.$

$f^{'}(x)=\dfrac{d[(x-a)^{m}(x-b)^{n}]}{dx}=(x-b)^{n}m(x-a)^{m-1}+(x-a)^{m}n(x-b)^{n-1}$

now putting $f^{'}(x)=0$ , we get ,

$=m(x-a)^{m-1}(x-b)^{n}+(x-a)^{m}n(x-b)^{n-1}=0$

$=(x-a)^{m-1}(x-b)^{n-1}[m(x-b)+n(x-a)]=0$

$(m+n)x=mb+na$

$x=\dfrac{(mb+na)}{m+n}$

$=\dfrac{(mb+na)}{m+n}>a$ , m and n are given that are positive numbers .

$=mb+na>ma+na\implies$ $m(b-a)>0$

Similarly ,

$=\dfrac{(mb+na)}{m+n}<b$

$=mb+na<mb+nb\implies n(a-b)<0$

So it implies that

$a<\dfrac{(mb+na)}{m+n}<b$

So , hence our rolle's theorem verifies.