Finding roots of the derivative of a polynomial.

[Jesssa]

hey

i'm trying to figure out how to approach part b of this problem,

http://imageshack.us/a/img850/6059/asdasdno.jpg

so i can see that you can apply the mean value theorem to p'(x)

so there exists some c between a and b such that

f'(c) = [f(b) - f(a)] / (b-a)=0

so p'(x) has a root between a and b,

however

P(x) = p'(x) + kp(x)

and p(x) only has roots a and b on the interval (a,b) and nothing else

so P(c) = 0 + kp(c) which is only 0 for k=0 or c=a or b


this is making me a big confused on how to continue with the question,

does anyone have any ideas?

 

[HallsofIvy]

An obvious first step is to differentiate the given function:
p'(x)= m(x-a)^{m-1}(x- b)^nq(x)+ n(x- a)^m(x- b)^{n-1}q(x)+ (x- a)^m(x- b)^n q'(x)
Now factor out (x- a)^{m-1} and (x- b)^{n-1}:
p'(x)= (x-a)^{m-1}(x- b)^{m-1}(m(x- b)^nq(x)+ n(x- a)q(x)+ (x- a)(x- b)q'(x))
so that r(x)= m(x- b)^nq(x)+ n(x- a)q(x)+ (x- a)(x- b)q'(x)

Now show that r(a) and r(b) have different signs.

 

[Jesssa]

thanks for replying!

sorry if i am miss understanding something, but it seems like you are describing the steps of part a of the question.

i'm having trouble seeing how the sign of r(a) and r(b) effects P(x) = p'(x) + kp(x) (part b)
since here the zero of P(x) is in the interval (a,b).

p'(x) is not 0 always at x=a or b because it will depend on the value of n and m,

and the fact that it says 'for any k in ℝ" makes me think that the kp(x) must be zero so
kp(x) = 0, but its for all k in ℝ so p(x) = 0 but that is only true at x=a or x=b
but if x=a or x=b then p'(x) is not equal to zero unless bot m,n≠1