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mathmath333
 one year ago
find
mathmath333
 one year ago
find

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\begin{align} &\text{if}\quad 3\leq x\leq 5 ,\quad x\in \mathbb{R} \hspace{.33em}\\~\\ &\text{find a and b such that } \hspace{.33em}\\~\\ &a\leq x^2 \leq b \end{align}}\)

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0i'm thinking of taking root as a first option

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, the minimum value of x^2 would be 0, which is fine since 3 <=x<=5. And the highest absolute value for 3<=x<=5 is 5, so then you would consider the maximum value of x^2 to be 25. So a = 0, b = 25.

nuttyliaczar
 one year ago
Best ResponseYou've already chosen the best response.1I agree with @Concentrationalizing that little trick you had there was subtle but clever ^^

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3the graph of \(y=x^2\) in the interval \([3, 5]\)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0what about this \(\large \color{black}{\begin{align} &\text{if}\quad 4\leq x^2\leq 25 ,\quad x\in \mathbb{R} \hspace{.33em}\\~\\ &\text{find a and b such that } \hspace{.33em}\\~\\ &a\leq x \leq b \end{align}}\)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0i didnt get ur graph

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0ok so he plotted \(y=x^2\) from \([2,5]\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3yes plotted \(y=x^2\) from \(y=4\) to \(y=25\)

phi
 one year ago
Best ResponseYou've already chosen the best response.1the problem with 0 to 25 is it allows x= 5 (for example)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0but i cant use graphing calculator , i need pen paper method

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\[4\leq x^2\leq 25 \] take squareroot through out \[\sqrt{4}\leq \sqrt{x^2}\leq \sqrt{25} \] which is same as \[2\leq x\leq 5 \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3yeah need to see \(\sqrt{x^2} = x\)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0so we can take square root in 'inequality' normally as in equations ?

phi
 one year ago
Best ResponseYou've already chosen the best response.1**so we can take square root in 'inequality' normally as in equations ?** if you make explicit assumptions about the sign of the root. e.g. take only the principal root

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\[x^2\ge a ~~\iff~~ x \ge \sqrt{a}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3assuming ofcourse \(a\gt 0\)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0how can i use wolfram to solve this problem

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3wolfram has no problem interpreting double sided inequalities http://www.wolframalpha.com/input/?i=solve+4%3C%3Dx%5E2%3C%3D25

phi
 one year ago
Best ResponseYou've already chosen the best response.1btw, for \[ {\begin{align} &\text{if}\quad 3\leq x\leq 5 ,\quad x\in \mathbb{R} \hspace{.33em}\\~\\ &\text{find a and b such that } \hspace{.33em}\\~\\ &a\leq x^2 \leq b \end{align}} \] it seems the best way to write the form is \[ 0 \le (x1)^2 \le 16 \] but that does use just x^2

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0how did u arrived from \(3\leq x\leq 5\) to \(0\leq (x1)^2\leq 16\)

phi
 one year ago
Best ResponseYou've already chosen the best response.1find the middle of the region 3 to 5 (at 1) x=1 is in the middle and we want the distance from the middle to be no more than 4 (i.e. (5  3)/2 = 8/2= 4) we the distance x1 <= 4 and that leads to the equation posted
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