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We need to find what $a$ is. Can I not solve this using the mean value theorem in the interval $(a,a+1)$?

I tried for a $c\in(a,a+1)$

$$f(c)=\int_a^{a+1}f(x)dx$$ but since all the roots lie in $(a,a+1), f(c)=0$ We could just integrate $f(x)$ and try different values of $a$ unless the equation is satisfied, but apparently this doesn't work.Why?

user26857
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Parth Shresth
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  • Could you elaborate how all the roots lie in $(a, a+1)$ implies $f(c) = 0$? – VTand Apr 19 '22 at 12:03
  • @VTand since $c\in(a,a+1)$ and $f(x)$ has all real roots in that interval can't we find a $c$ such that $f(c)=0$? – Parth Shresth Apr 19 '22 at 12:06
  • Have you attempted to draw the graphical representation of $y=2x^5+5x^4+...$ in order to have an idea of the value of $a$ ? – Jean Marie Apr 19 '22 at 12:07
  • Using the characterisation $f(c)=\int_a^{a+1}...$ is a little "overkill" IMHO... – Jean Marie Apr 19 '22 at 12:09
  • @JeanMarie Yeah, I did analyse it's derivative and found that the function is strictly increasing and used the intermediate value to "locate" the root but I was wondering why this doesn't work. – Parth Shresth Apr 19 '22 at 12:10
  • @JeanMarie well, this was the thing that popped in my mind furst, figuring out that the derivative was positive took me a bit too long – Parth Shresth Apr 19 '22 at 12:11
  • @ParthShresth The value $c$ is already fixed when you define $f(c) = \int_a^{a+1} f(x) : dx$ by MVT. – VTand Apr 19 '22 at 12:13
  • What can you say about the sign of your function from one end of the interval to the other, and how can you use this to solve your problem? – Mark Bennet Apr 19 '22 at 12:16
  • @MarkBennet the signs of the function at the ends of the interval must be opposite, which suggest that $f(a)f(a+1)<0$ but that doesnt seem to help – Parth Shresth Apr 19 '22 at 12:19
  • Well, on how many distinct intervals between successive integers can the function change sign? – Mark Bennet Apr 19 '22 at 12:24
  • @MarkBennet It should be atleast one as the function has an odd leading power but how should I proceed – Parth Shresth Apr 19 '22 at 12:31
  • Using the Cauchy bound for polynomial roots, you can see that any root must be on the interval $(-6, -\frac 16)$. Looking for the roots of the derivative in the same way, you can narrow it down to the interval $-4, -\frac 14)$. This does not leave many possible values for $a$. – PierreCarre Apr 19 '22 at 12:36
  • Please give the full problem statement in the body of your Question, not relying on the title for that purpose. The jump from title to body and the title's relative brevity can make reading difficult. – hardmath Apr 19 '22 at 12:44
  • @PierreCarre I didn't know a cauchy bound was. I googled it and found out that it gives only an upper bound, do you mind elaborating how did you find a lower bound for the roots? – Parth Shresth Apr 19 '22 at 12:46
  • But count the roots - how many intervals can have a change of sign? Then it is just a search for an interval, and there are various ways reducing the search space. – Mark Bennet Apr 19 '22 at 12:47
  • plot 2x^5+5x^4+10x^3+10x^2+10x+10 in Wolfram Alpha shows you what $a$ must be, and you can verify it easily by evaluating $f(a)$ and $f(a+1)$. – TonyK Apr 19 '22 at 13:02

1 Answers1

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Let $f$ be the polynomial. Since $f$'s degree is odd, there is at least one root. This combined with the fact that the polynomial is strictly increasing, gives you that there is exactly one root. Now, we evaluate $f$ for some easy negative numbers (clearly the roots must be negative): $$f(-1)=3\text{ and }f(-2)=-34.$$ Therefore, the root(s) lies in $(-2,-1)$.

moqui
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