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January 28th, 2015, 11:41 PM   #1
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Question 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.
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January 29th, 2015, 02:10 AM   #2
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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.
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January 29th, 2015, 02:51 AM   #3
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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?
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January 29th, 2015, 04:07 AM   #4
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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 $\frac14$.
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January 30th, 2015, 05:23 AM   #5
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thank u v8archie

It help me a lot

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