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 January 28th, 2015, 11:41 PM #1 Newbie   Joined: Jan 2015 From: IRAN Posts: 11 Thanks: 0 what's the minimum of this function? Hi everyone can anybody help me to solve this function? plz solve it without plot and with shortest solution... min f(x) ? f(x)=Max{x^2,|x-3/4|} thanks a lot... Last edited by mili; January 29th, 2015 at 12:22 AM.
 January 29th, 2015, 02:10 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,302 Thanks: 1688 What exactly does "without plot" mean? Explaining the answer requires what's effectively a description of the plot even if the function's graph isn't sketched.
 January 29th, 2015, 02:51 AM #3 Newbie   Joined: Jan 2015 From: IRAN Posts: 11 Thanks: 0 I asked this question from four person and all of them siad just one solution... and that solution is this graph abs(x-3/4) and x^2 - Wolfram|Alpha I wanna solve it without drawing any plot but one person solve it without plot but that solution is so long... may anybody help me?
 January 29th, 2015, 04:07 AM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,346 Thanks: 2466 Math Focus: Mainly analysis and algebra First you must determine the value of the function everywhere. To do this we examine$$x^2 \gt |x - \tfrac34|$$ For $x \ge \tfrac34$ we have \begin{aligned}x^2 &\gt x - \tfrac34 \\ x^2 - x + \tfrac34 &\gt 0\end{aligned} The discriminant of the left side is $1 - 3 = -2 \lt 0$, so the inequality is always true or always false. Putting $x=1$ shows the truth of the inequality in this case and thus everywhere. So, for $x \ge \tfrac34$ the function is equal to $x^2$ and on this domain, the minimum is $(\frac34)^2 = \frac9{16}$. Now we look at $x \le \tfrac34$. Here we seek \begin{aligned}x^2 &\gt \tfrac34 - x \\ x^2 + x - \tfrac34 &\gt 0 \\ (x + \tfrac32)(x - \tfrac12) &\gt 0\end{aligned} So for $x \le -\tfrac32$ the inequality is true and our function is equal to $x^2$. Its minimum on this domain is $(-\frac32)^2 = \frac94$. Similarly, for $x \ge \frac12$ our function is equal to $x^2$. Its minimum on this domain is $(\frac12)^2 = \frac14$. On the rest of the real domain, ($-\frac32 \le x \le \tfrac12$) our function is linear ($\tfrac34 - x$) and this line must join two parts of the graph. So the function's minimum is least of the minima we have found. Thus the minimum is $\frac14$. Thanks from mili
 January 30th, 2015, 05:23 AM #5 Newbie   Joined: Jan 2015 From: IRAN Posts: 11 Thanks: 0 thank u v8archie It help me a lot ♥♡♥

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