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Let $$f(x) = (x-1)(x-2)(x-3)(x-4)(x-5),\quad -\infty< x<\infty.$$ The number of distinct roots of equation $$\frac{d}{dx}f(x) = 0$$ is exactly ?

Source.

The only method that I know is to multiply and then find the derivative of function then apply Sturm's theorem but it seems vague when you have to solve the question in 3 to 5 minutes . So you are requested to suggest a plausible alternative approach.

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See also the comments here. – Dietrich Burde Jan 4 at 14:29
    
@DietrichBurde but my problem is , I am facing lots of tricky questions daily & they are saying that " There are many approaches to give practical short cuts in relation to the polynomials you might find in an exam . " But I am unable to find shortcut for this question . I am doing other question by suggested methods but I have posted this question in search of a short-cut approach . – Tesla Jan 4 at 14:38
up vote 8 down vote accepted

$f(x)$ is a polynomial of degree 5, $f'(x)$ is one of degree 4. So $f'$ has 4 zeros. As the zeros of $f$ are distinct, it has no zeros in common with $f'$. By the mean value theorem, there has to be a zero of $f'$ between each consecutive pair of zeros of $f$, i.e., there are at least 4 different zeros of $f'$ (there might be several between consecutive zeros, or ones outside the range of zeros of $f$). But as $f'$ is a quartic, there are exactly 4.

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The multiplicity of each of the roots of $f(x)$ is $1$, which means that the graph of the function of $f(x)$ crosses the $x$-axis at those roots. By drawing a sketch, it becomes obvious that the derivative must have precisely four real roots.

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+1 I think drawing the sketch is the fastest way to see this with the least amount of knowledge of analysis required. It does, however, require one to be able to sketch polynomials, which is sadly not taught as widely as it should be (i.e., universally!) If a graphing calculator is allowed to anyone presented with such a challenge, hopefully they at least know how to use that to quickly graph a polynomial and come up with the answer. – Todd Wilcox Jan 4 at 18:45
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+1 I don't think you really even need to sketch. You know polynomials are continuous and differentiable everywhere. You also know that at each of the 5 zeros the sign of the polynomial changes and none of them are multiple (so you can't get zero derivatives at them). This should make it possible to visualize what it looks like. – DRF Jan 5 at 8:17

Here's an alternative to using Rolle's Theorem, which can also be done in a couple of minutes.

Let $x-3=u$. Then the polynomial becomes

$$u(u^2-1)(u^2-4)=u(u^4-5u^2+4)=u^5-5u^3+4u$$

so the derivative is

$$5u^4-15u^2+4$$

which is quadratic in $u^2$. You can do is either using the quadratic formula,

$$U=u^2={15\pm\sqrt{225-80}\over10}$$

so simply by noting that $5U^2-15U+4$ is clearly positive for $U\le0$ and negative at $U=1$, so it has two distinct positive roots, hence the quartic derivative has $4$ real roots, all distinct.

It took me more than five minutes to write this up, but the scratchwork and thinking took less than three. The key was to notice that a simple shift makes it easy to expand the factored quintic.

Remark: This approach clearly only works because the roots of the quintic are equally spaced. If the problem started, say with $(x-1)(x-2)(x-4)(x-8)(x-16)$, then Rolle's Theorem would be your best bet. Still, it's a useful alternative to have on tap; suppose, for example, the question had asked for the number of solutions to $f'(x)=1$. Try answering that armed only with Rolle!

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Think about the graph of the function: it will have zeroes at $x = 1, 2, 3, 4, 5$, and will be nonzero in between (since $f(x) = 0$ requires that one of the factors of $(x - 1)(x - 2)(x - 3)(x - 4)(x - 5)$ is zero). So $f$ will have at least one local extremum on each of the intervals $(1, 2), (2, 3), (3, 4)$ and $(4, 5)$ (one can see this intuitively and make it rigorous by using the extreme value theorem). One can make this rigorous by using the mean value theorem.

Since the degree of $f'(x)$ will be four (which can be seen by writing $f(x)$ as a polynomial and differentiating it), it won't have more than four zeroes. So $f'$ must have exactly four zeroes.

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