Show that the derivative of f(x) = (x-a)m (x-b)n vanishes at some point between a & b

  • #1
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.
 

Answers and Replies

  • #2
2
0


easily can be shown by rolle's theorem
 
  • #3


We cant use that yet :(
 
  • #4
323
56


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?
 
  • #5
HallsofIvy
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Since this question has nothing to do with "Differential Equations", I am moving it to Calculus and Beyond Homework.
 
  • #6
HallsofIvy
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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.
 
  • #7


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

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