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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?

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  • $\begingroup$ Could you elaborate how all the roots lie in $(a, a+1)$ implies $f(c) = 0$? $\endgroup$
    – VTand
    Apr 19 at 12:03
  • $\begingroup$ @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 $f(c)=0$? $\endgroup$
    – Parth Shresth
    Apr 19 at 12:06
  • $\begingroup$ Have you attempted to draw the graphical representation of $y=2x^5+5x^4+...$ in order to have an idea of the value of $a$ ? $\endgroup$
    – Jean Marie
    Apr 19 at 12:07
  • $\begingroup$ Using the characterisation $f(c)=\int_a^{a+1}...$ is a little "overkill" IMHO... $\endgroup$
    – Jean Marie
    Apr 19 at 12:09
  • $\begingroup$ @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. $\endgroup$
    – Parth Shresth
    Apr 19 at 12:10

1 Answer 1

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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 $(-2,-1)$.

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