## mathmath333 one year ago find

1. mathmath333

\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

b is not 5

3. xapproachesinfinity

hmm

4. anonymous

5. xapproachesinfinity

i'm thinking of taking root as a first option

6. anonymous

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

I agree with @Concentrationalizing that little trick you had there was subtle but clever ^^

8. welshfella

yes

9. ganeshie8

the graph of $$y=x^2$$ in the interval $$[-3, 5]$$

10. mathmath333

ok thnx

11. mathmath333

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

|dw:1433705362979:dw|

13. ybarrap

*Not to scale :)

14. mathmath333

i didnt get ur graph

15. mathmath333

ok so he plotted $$y=x^2$$ from $$[2,5]$$

16. mathmath333

*from $$[4,25]$$

17. ganeshie8

yes plotted $$y=x^2$$ from $$y=4$$ to $$y=25$$

18. phi

the problem with 0 to 25 is it allows x= -5 (for example)

19. mathmath333

but i cant use graphing calculator , i need pen paper method

20. ganeshie8

$4\leq x^2\leq 25$ take squareroot through out $\sqrt{4}\leq \sqrt{x^2}\leq \sqrt{25}$ which is same as $2\leq |x|\leq 5$

21. ganeshie8

yeah need to see $$\sqrt{x^2} = |x|$$

22. mathmath333

so we can take square root in 'inequality' normally as in equations ?

23. 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

24. ganeshie8

$x^2\ge a ~~\iff~~ |x| \ge \sqrt{a}$

25. ganeshie8

assuming ofcourse $$a\gt 0$$

26. mathmath333

how can i use wolfram to solve this problem

27. ganeshie8

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

28. 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

29. phi

*not

30. mathmath333

how did u arrived from $$-3\leq x\leq 5$$ to $$0\leq (x-1)^2\leq 16$$

31. 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