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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: 20,973 Thanks: 2223 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: 7,685 Thanks: 2665 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 ♥♡♥ Tags function, minimum Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post condemath2 Calculus 1 June 6th, 2014 05:13 PM MATHEMATICIAN Calculus 9 August 31st, 2013 10:33 AM david hilbert Complex Analysis 3 January 28th, 2013 02:33 AM amateurmathlover Calculus 4 April 22nd, 2012 12:39 AM bilgisiz Number Theory 1 March 8th, 2011 02:27 PM

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