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Obtain the number of real roots of the following quintic polynomial $$f(x)= x^5+x^3-2x+1$$

My approach:

First, on seeing the polynomial, it should be clear that it is continuous. Also, $f(-\infty)=-\infty$ and $f(\infty)=\infty$. Hence, the function has at least one root. Now, the derivative is calculated as:

$$f'(x)=5x^4+3x^2-2$$

which gives the two solutions as $\pm\sqrt{\frac25}$. Now, we need to test for the roots in each of the intervals. In the first Interval, the value of function is negative at $-\infty$ and positive at $-\sqrt{\frac25}$. Hence, function will have one root in this interval. Then, in the second Interval function decreases but never touches $x$-axis as it takes positive value at $\sqrt{\frac25}$. Also, the function again start to increase in the third Interval and is positive at $\infty$. Hence, it has only one root.

Is my approach correct? My textbook gives the answer 3 using Rolle's theorem. It says that since the $f'(x)$ has two roots, then $f(x)$ will have three roots using Rolle's theorem. Thanks.

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    $\begingroup$ According to WA , there is only one real root: your book is wrong. $\endgroup$
    – Crostul
    Feb 1, 2018 at 8:45
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    $\begingroup$ Since $f'(x)$ has two roots, by Rolle's theorem $f(x)$ cannot have more than three roots. But whether the answer is one, two or three roots the theorem cannot tell you. $\endgroup$
    – Arthur
    Feb 1, 2018 at 11:13
  • $\begingroup$ @Arthur, thank you so much. $\endgroup$
    – userNoOne
    Feb 1, 2018 at 11:39

2 Answers 2

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Your approach is OK. Here is just another way.

By Descartes' rule of signs, $f(-x) = -x^5-x^3+2x+1$ has exactly one sign change, so there is exactly one negative real root.

As $f(x) = x^5+x^3-2x+1$ itself has two sign changes, we can have either no or two positive roots. We are left to show there is none, which can be observed by the following AM-GM: $$f(x) = x^5 + x^3 + 6\times \tfrac16 - 2x \geqslant \frac8{\sqrt[8]{6^6}}x -2x > 0$$ as $8 > 2\cdot 6^{3/4} \iff 4^4 > 6^3 \iff 256 > 216$ which is true.

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    $\begingroup$ Your method is unique. Nice. Thank you so much. $\endgroup$
    – userNoOne
    Feb 1, 2018 at 9:08
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Your approach is correct, and the textbook misses the point. There is no direct relation between the number of zeros of $f'(x)$ and the number of zeros of $f(x)$, since derivatives ignore constant terms and (dually) integrals are only defined up to a constant.

(The number of zeros of $f'(x)$ plus one is an upper bound for the number of roots of $f(x)$, but the latter may go all the way down to zero, as pointed out in the comments on this answer.)

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  • $\begingroup$ Thank you so much. I was really confused in this. $\endgroup$
    – userNoOne
    Feb 1, 2018 at 9:00
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    $\begingroup$ There is a relation between the number of roots of the derivative and the upper bound of the number of roots of the original function, but not between the derivative and the lower bound. $\endgroup$
    – Acccumulation
    Feb 1, 2018 at 17:33

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