Question

Verify Rolle's Theorem for furetion
f(x) = (x-a)^m (x-b)^n in the
interval [a, b] where m and n
are positive integers​

Answer 1

Answer:

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

Step-by-step explanation:

Given that

$f(x)=(x-a)^m(x-b)^n$

We have to verify Rolle's theorem in interval [a,b]

we know that in  Rolle's theorem ${f(c)}'=0$

  C∈[a,b]

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

Now lets take $(x-a)^m(x-b)^n$ common from above the equation

${f(x)}'=(x-a)^m(x-b)^n\left( \frac{m}{x-a}+\frac{n}{x-b}\right)$

So now by equating

${f(x)}'=0$

${f(x)}'=(x-a)^m(x-b)^n\left( \frac{m}{x-a}+\frac{n}{x-b}\right)=0$

So

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

Lets take values of of a,b,n and m to verify above equation

a=1,b=2,m=3,n=4

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

$x=\dfrac{3\times 2+4\times 1}{3+4}$

So

$x=\dfrac{10}{7}$

It means that 10/7 lies in between [1,2].So it means it verify Rolle's theorem.