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January 29th, 2015, 12:41 AM  #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,x3/4} thanks a lot... Last edited by mili; January 29th, 2015 at 01:22 AM. 
January 29th, 2015, 03:10 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,267 Thanks: 1958 
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, 03:51 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(x3/4) and x^2  WolframAlpha 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, 05:07 AM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,599 Thanks: 2586 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 $\frac14$. 
January 30th, 2015, 06:23 AM  #5 
Newbie Joined: Jan 2015 From: IRAN Posts: 11 Thanks: 0 
thank u v8archie It help me a lot ♥♡♥ 

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