- #1

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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.

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- Thread starter tachyon_man
- Start date

- #1

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My attempt:

f(x) = (x-a)

f '(x) = m(x-a)

f '(x) = [(x-a)

And this is as far as I got.

- #2

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easily can be shown by rolle's theorem

- #3

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We cant use that yet :(

- #4

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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

Science Advisor

Homework Helper

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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

Science Advisor

Homework Helper

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Set that equal to 0 and solve for x:

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.

[(x-a)

Since we are looking at x

- #7

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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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