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If $a, b, p$ and $q$ are positive with $a<b$ find the x-coordinate of the stationary point of the curve $y=(x-a)^p(x-b)^q$ in the domain $a<x<b$.

This question is from the chain rule chapter, so it doesn't involve the product or quotient rule. How do I do this question?

Jim
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  • With the chain and product rules. – Zook May 06 '14 at 18:24
  • you should probably use the chain rule. Use it to calculate the derivatives of $(x-a)^p$ and $(x-b)^q$ and then use the product rule to calculate the derivate of $(x-a)^p\cdot (x-b)^q$ using the derivates of both factors. – dinosaur May 06 '14 at 18:24
  • Does the chain rule chapter by any chance come after the product rule chapter...? – Zook May 06 '14 at 18:25
  • Nope, it comes before, which is why I am convinced there is a way to do it with the chain rule. – Jim May 06 '14 at 18:26
  • @Zook any advances? – Jim May 06 '14 at 18:44
  • I don't see any way to do it without the product rule unless you know the values of $p$ and $q$, in which case you can expand the polynomials. Or if $p=q$, in which case you can write it as $y=[(x-a)(x-b)]^p$ and expand the inner quadratic. – Zook May 06 '14 at 21:11
  • Textbooks have been known to make mistakes... – Zook May 06 '14 at 21:17
  • The answer, using the product rule, is actually pretty cool, it's a weighted average of $a$ and $b$ depending on the values of the exponents. – Zook May 06 '14 at 21:19
  • You could do some really really nasty expansion and factoring with the binomial theorem... I could reverse engineer it by getting there ahead of time with the product rule if you want... – Zook May 06 '14 at 21:24

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