Let $f(x)=y$ and $\{x\}=a$
$$(y+\frac 34)^2 =2a-a^2$$
$$y^2+\frac 94 +\frac{3y}{2} =2a-a^2$$
If I had a singular $\{x\}$ term I could have simply applied the inequality $0\le \{x\}<1$ But instead I have a polynomial. What should I do in this case?
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$\begingroup$ @PeterForeman I don’t think AM-GM can be applied, and the double derivative of $2a-a^2$ shows it’s decreasing (my knowledge on calculus is currently limited) . The function will be have max value for min $x$. But on putting $a=0$ and $a=1$ we get $1<a(2-a)\le 0$ which is obviously wrong $\endgroup$– AdityaJun 23, 2020 at 8:44
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$\begingroup$ Please verify this $a-1\in [-1,0) \implies (a-1)^2 \in [1,0) \implies a^2-2a\in [0,-1) \implies 2a-a^2 \in [0,1)$ $\endgroup$– AdityaJun 23, 2020 at 8:51
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$\begingroup$ @PeterForeman yes it was a typo. So now I should just square root the interval and subtract $\frac 34$ from it? $\endgroup$– AdityaJun 23, 2020 at 8:59
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$\begingroup$ Yes that should work. I would note that you cannot square root an interval but you can apply the fact that if $x\in[a,b]$ with $0\lt a\lt b$ we have $\sqrt{x}\in[\sqrt{a},\sqrt{b}]$. $\endgroup$– Peter ForemanJun 23, 2020 at 9:02
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$\begingroup$ @PeterForeman just wanted to confirm, $\{,\}$ is the fractional part function $\endgroup$– AdityaJun 23, 2020 at 11:22
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3 Answers
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Hint.
Your function is a bit particular : the $\{x\}$ makes it a bit more difficult to reason about.
But in fact you can just write $f(x) = g(\{x\})$ for a certain g. and since $\{x\}$ spans over $[0;1]$ when $x$ spans over $\mathbb{R}$, you just have to study g
(Just draw it: since it is a 2nd degree polynomial, it is a parabola, and its range will appear clear on the drawing. If you don’t like drawing, just differentiate it)
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$\begingroup$ Well basically you need to change your variable. Then you only have to study $g(a) = \sqrt{2 a - a^2} - 3/4$. In fact you only have to study the range of $h(a) = 2*a - a^2$ since $y -> \sqrt{y} - 3/4$ is clearly monotonic. h is a parabola, so finding its range should be easy (just draw it to understand) $\endgroup$– NephanthJun 23, 2020 at 9:01
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Hint:
$$2\{x\}(1-\{x\})$$ is non-negative and peaks at $\frac12$ (parabola).
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$\begingroup$ I got the parabola in the form of $(x-1)^2=-(y-1)$. How is it non negative (i know it’s non negative by looking at it, but the parabola says it isn’t) $\endgroup$– AdityaJun 23, 2020 at 8:48
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$\begingroup$ I didn't say that "the parabola is non-negative", did I ? :-) $\endgroup$– user65203Jun 23, 2020 at 8:52
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If you differentiate the function, you get $$1-2x = 0 \text { or } x = \frac{1}{2}$$ $$\text {Also given that } \sqrt{2x-2x^2} \text { can only be a real number if } 2x-x^2 \ge 0,$$ $$\text {The min value of } (2x-2x^2) = 0 \text { , or lower and upper bound of } x = 0,1$$ $$\text {So, } 0 \le x \le 1 \text { with maxima at } x = \frac {1} {2}$$
It is a restricted parabola.
answered Jun 23, 2020 at 11:03
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$\begingroup$ $x$ is inside the fractional part function though $\endgroup$– AdityaJun 23, 2020 at 11:22