Derivative of f(x) Vanishing Between a & b for Positive Integer m & n

[tachyon_man]

Question: Show that the derivative of f(x) = (x-a)m (x-b)n vanishes at some point between a and b if m and n are positive integers.
My attempt:
f(x) = (x-a)^m (x-b)ⁿ
f '(x) = m(x-a)^m-1 (x-b)ⁿ + n(x-a)^m (x-b)^n-1
f '(x) = [(x-a)^n-1 (x-b)^n-1 ] [(m)(x-b) +(n)(x-a)]
And this is as far as I got.

 

[anubhab92]

easily can be shown by rolle's theorem

 

[tachyon_man]

We can't use that yet :(

 

[alan2]

You have the expression for the derivative. Set it equal to zero and you will find one value of x for which the derivative is always zero, namely x=(mb+na)/(m+n). You then have to show that it is always between a and b if m and n are positive integers. Actually, m and n can be any reals greater than or equal to 1.

Why can't you use Rolle's theorem if you understand it?

 

[HallsofIvy]

Since this question has nothing to do with "Differential Equations", I am moving it to Calculus and Beyond Homework.

 

[HallsofIvy]

Question: Show that the derivative of f(x) = (x-a)m (x-b)n vanishes at some point between a and b if m and n are positive integers.
My attempt:
f(x) = (x-a)^m (x-b)ⁿ
f '(x) = m(x-a)^m-1 (x-b)ⁿ + n(x-a)^m (x-b)^n-1
f '(x) = [(x-a)^n-1 (x-b)^n-1 ] [(m)(x-b) +(n)(x-a)]
And this is as far as I got.

Set that equal to 0 and solve for x:
[(x-a)^n-1 (x-b)^n-1 ] [(m)(x-b) +(n)(x-a)]= 0
Since we are looking at x between a and b, x- a and x- b are not 0 so we can divide by them. That leaves m(x- b)+ n(x- a)= mx- mb+ nx- na= (m-n)x- mb-na= 0. Solve that for x and show that it lies between a and b.

 

[tachyon_man]

Thanks guys! I ended up using Rolles theorem. I'm still waiting to see if he accepts it as a valid approach.