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If $f(x) $ Is a continuous function $ \forall\ x \in R$ and satisfies $x^2+x \{f(x)\} - 3 = \sqrt{3} \ f(x) \ \forall\ x \in R$

Find $f(\sqrt{3})$.

$\{ t\}$ is the fractional part of $t$.

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My attempt: I substituted $\sqrt{3}$ in that and could only conclude that $0 \leq f(\sqrt{3}) < 1$

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    $\begingroup$ @JonatanB.Bastos, if $0\le a\lt1$ then the integral part of $a$ is $0$, but the fractional part is $a$. $\endgroup$ – Barry Cipra Dec 29 '17 at 13:37
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    $\begingroup$ I think that the fractional part is anything to the right of the decimal not the integer part $\endgroup$ – Satish Ramanathan Dec 29 '17 at 13:37
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    $\begingroup$ @SatishRamanathan, not that it matters to the problem here, but what would you say is the fractional part of $-0.25$? $\endgroup$ – Barry Cipra Dec 29 '17 at 13:42
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    $\begingroup$ Compute $f(0)=-\sqrt{3}$. Observe that $f$ is unbounded from above, since $x^2\to+\infty$ as $x\to+\infty$. Therefore it has some zero for $x>0$. Let $x_0$ be such that $f(x_0)=0$. Then $x_0^2-3=0$. Therefore $x_0=\sqrt{3}$. So, $f(\sqrt{3})=0$. $\endgroup$ – user515219 Dec 29 '17 at 13:43
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    $\begingroup$ @despaigne, I believe you should add your comment as another answer, as it illustrates a very interesting way to approach the problem. A question: do you think we can make your approach work if we assume the equation to hold only in a neighborhood of $\sqrt{3}$ (and not for every $x \in \mathbb{R})$? $\endgroup$ – Pedro M. Dec 29 '17 at 13:59
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By substituting $\sqrt{3}$ in there, you get that $f(x) - \{f(x)\} = 0$, which implies that $0 \leq a < 1$ as you concluded.

Now suppose $f(\sqrt{3}) = a, 0 < a < 1$. Then, by continuity, $f$ satisfies the equation $x^2 + xf(x) - 3 = \sqrt{3} f(x)$ in a neighborhood of $x = \sqrt{3}$, so that $$f(x) = \frac{x^2 - 3}{\sqrt{3} - x} = -(\sqrt{3} + x),$$

which implies that $f(\sqrt{3}) = - 2\sqrt{3} < 0$ (a contradiction because $a > 0$). Therefore, we must have $f(\sqrt{3}) =0$.

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  • $\begingroup$ @samjoe: Sorry, you are correct. The signs got scrambled in my head. I have edited the answer to reflect this. $\endgroup$ – Pedro M. Dec 29 '17 at 13:49
  • $\begingroup$ No problem just a typo! But can you tell me how we conclude $f(\sqrt 3) = 0$ in the end? $\endgroup$ – jonsno Dec 29 '17 at 13:51
  • $\begingroup$ We know that $0 \leq f(\sqrt{3}) < 1$, and we concluded that we cannot have $f(\sqrt{3}) > 0$ (because we reach a contradiction). Therefore we must have $f(\sqrt{3}) = 0$. $\endgroup$ – Pedro M. Dec 29 '17 at 13:52
  • $\begingroup$ Note that the argument that $x^2 - xf(x) - 3 = \sqrt{3} f(x)$ in a neighborhood of $\sqrt{3}$ only works if $f(\sqrt{3})$ is strictly greater than zero (so that it is strictly greater than zero in a neighborhood as well). $\endgroup$ – Pedro M. Dec 29 '17 at 13:54
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    $\begingroup$ @Rick Only at $x = \sqrt{3}$, and not in a neighborhood of $\sqrt{3}$, and this is the key point. We may have $f(x) < 0$ for some $x$ near $\sqrt{3}$ and the equation does not hold. This is the key point (neighborhood), otherwise the step where we "divide by $\sqrt{3} - x$" doesn't make sense. $\endgroup$ – Pedro M. Dec 29 '17 at 14:46

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