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mathmath333

  • one year ago

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} &\text{if}\quad -3\leq x\leq 5 ,\quad x\in \mathbb{R} \hspace{.33em}\\~\\ &\text{find a and b such that } \hspace{.33em}\\~\\ &a\leq x^2 \leq b \end{align}}\)

  2. Zarkon
    • one year ago
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    b is not 5

  3. xapproachesinfinity
    • one year ago
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    hmm

  4. anonymous
    • one year ago
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    SOrry, 25, my bad

  5. xapproachesinfinity
    • one year ago
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    i'm thinking of taking root as a first option

  6. anonymous
    • one year ago
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    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.

  7. nuttyliaczar
    • one year ago
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    I agree with @Concentrationalizing that little trick you had there was subtle but clever ^^

  8. welshfella
    • one year ago
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    yes

  9. ganeshie8
    • one year ago
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    the graph of \(y=x^2\) in the interval \([-3, 5]\)

  10. mathmath333
    • one year ago
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    ok thnx

  11. mathmath333
    • one year ago
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    what about this \(\large \color{black}{\begin{align} &\text{if}\quad 4\leq x^2\leq 25 ,\quad x\in \mathbb{R} \hspace{.33em}\\~\\ &\text{find a and b such that } \hspace{.33em}\\~\\ &a\leq x \leq b \end{align}}\)

  12. ybarrap
    • one year ago
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    |dw:1433705362979:dw|

  13. ybarrap
    • one year ago
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    *Not to scale :)

  14. mathmath333
    • one year ago
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    i didnt get ur graph

  15. mathmath333
    • one year ago
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    ok so he plotted \(y=x^2\) from \([2,5]\)

  16. mathmath333
    • one year ago
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    *from \([4,25]\)

  17. ganeshie8
    • one year ago
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    yes plotted \(y=x^2\) from \(y=4\) to \(y=25\)

  18. phi
    • one year ago
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    the problem with 0 to 25 is it allows x= -5 (for example)

  19. mathmath333
    • one year ago
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    but i cant use graphing calculator , i need pen paper method

  20. ganeshie8
    • one year ago
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    \[4\leq x^2\leq 25 \] take squareroot through out \[\sqrt{4}\leq \sqrt{x^2}\leq \sqrt{25} \] which is same as \[2\leq |x|\leq 5 \]

  21. ganeshie8
    • one year ago
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    yeah need to see \(\sqrt{x^2} = |x|\)

  22. mathmath333
    • one year ago
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    so we can take square root in 'inequality' normally as in equations ?

  23. phi
    • one year ago
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    **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

  24. ganeshie8
    • one year ago
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    \[x^2\ge a ~~\iff~~ |x| \ge \sqrt{a}\]

  25. ganeshie8
    • one year ago
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    assuming ofcourse \(a\gt 0\)

  26. mathmath333
    • one year ago
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    how can i use wolfram to solve this problem

  27. ganeshie8
    • one year ago
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    wolfram has no problem interpreting double sided inequalities http://www.wolframalpha.com/input/?i=solve+4%3C%3Dx%5E2%3C%3D25

  28. phi
    • one year ago
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    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

  29. phi
    • one year ago
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    *not

  30. mathmath333
    • one year ago
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    how did u arrived from \(-3\leq x\leq 5\) to \(0\leq (x-1)^2\leq 16\)

  31. phi
    • one year ago
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    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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