Let $a, b, c$ be distinct real numbers. Then find the number of real solutions of
$(x − a)^5 + (x − b)^3 + (x − c)$
I can't understand how there will be any solution. The polynomial is not equated with anything.
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2Do you mean "Then find the number of real solutions of ..."? – Thomas Andrews May 13 '14 at 11:47
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Sometimes polynomials' roots are called solutions. Hence I believe Awesome's answer is correct. See here: http://mathworld.wolfram.com/PolynomialRoots.html – rubik May 13 '14 at 12:23
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@ThomasAndrews Yes. I have edited it. – Rudstar May 13 '14 at 13:48
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In my understanding, there are roots of a polynomial (function) and solutions of a polynomial equations. So roots would be correct. – Lutz Lehmann May 15 '14 at 08:53
1 Answers
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Assuming you equated it to $0$,
$$\lim_{x\to -\infty}f(x)=-\infty \text{ and } \lim_{x\to \infty}f(x)=\infty$$
$$f'(x)=5(x-a)^4+3(x-b)^2+1>0\implies \text{Exactly one real root }$$
This is because the function is strictly increasing. It shall cross $x$-axis only one time. Therefore, only one solution is possible. Perhaps a graph would help.
It is highly likely that this is the case.
evil999man
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2It doesn't matter what it's equated to, there'll be just one solution. – Barry Cipra May 13 '14 at 11:56
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@Awesome: I think that here solution is used as a synonym of root, as opposed to an equation's solution. – rubik May 13 '14 at 12:26
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@rubik Yes. This seems to be the case. Perhaps if answer is available to OP he can verify. – evil999man May 13 '14 at 12:27
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@Awesome I did not get how $f'(x)\implies$ exactly one real root. Could you please explain? – Rudstar May 13 '14 at 13:51
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1This is Rolle's theorem, if there are two roots, the derivative is zero at least once in between. – Lutz Lehmann May 15 '14 at 08:54