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\(\large \color{black}{\begin{align} &\text{if}\quad -3\leq x\leq 5 ,\quad x\in \mathbb{R} \hspace{.33em}\\~\\ &\text{find a and b such that } \hspace{.33em}\\~\\ &a\leq x^2 \leq b \end{align}}\)
b is not 5
hmm

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Other answers:

SOrry, 25, my bad
i'm thinking of taking root as a first option
Well, 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.
I agree with @Concentrationalizing that little trick you had there was subtle but clever ^^
yes
the graph of \(y=x^2\) in the interval \([-3, 5]\)
ok thnx
what about this \(\large \color{black}{\begin{align} &\text{if}\quad 4\leq x^2\leq 25 ,\quad x\in \mathbb{R} \hspace{.33em}\\~\\ &\text{find a and b such that } \hspace{.33em}\\~\\ &a\leq x \leq b \end{align}}\)
|dw:1433705362979:dw|
*Not to scale :)
i didnt get ur graph
ok so he plotted \(y=x^2\) from \([2,5]\)
*from \([4,25]\)
yes plotted \(y=x^2\) from \(y=4\) to \(y=25\)
  • phi
the problem with 0 to 25 is it allows x= -5 (for example)
but i cant use graphing calculator , i need pen paper method
\[4\leq x^2\leq 25 \] take squareroot through out \[\sqrt{4}\leq \sqrt{x^2}\leq \sqrt{25} \] which is same as \[2\leq |x|\leq 5 \]
yeah need to see \(\sqrt{x^2} = |x|\)
so we can take square root in 'inequality' normally as in equations ?
  • phi
**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
\[x^2\ge a ~~\iff~~ |x| \ge \sqrt{a}\]
assuming ofcourse \(a\gt 0\)
how can i use wolfram to solve this problem
wolfram has no problem interpreting double sided inequalities http://www.wolframalpha.com/input/?i=solve+4%3C%3Dx%5E2%3C%3D25
  • phi
btw, 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 \[ 0 \le (x-1)^2 \le 16 \] but that does use just x^2
  • phi
*not
how did u arrived from \(-3\leq x\leq 5\) to \(0\leq (x-1)^2\leq 16\)
  • phi
find 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 |x-1| <= 4 and that leads to the equation posted

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