## anonymous one year ago If f(x)=sqrt2-x and g(x) = x − 1, then find the domain of h(x) = f(x) g(x).

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1. jim_thompson5910

you wrote h(x) = f(x) g(x) but there's a gap between f(x) and g(x). Is there a symbol missing?

2. anonymous

division symbol

3. jim_thompson5910

is f(x) this? $\Large f(x) = \sqrt{2-x}$ OR is it this? $\Large f(x) = \sqrt{2}-x$

4. anonymous

the first

5. jim_thompson5910

so h(x) would be $\Large h(x) = \frac{f(x)}{g(x)}$ $\Large h(x) = \frac{\sqrt{2-x}}{x-1}$

6. jim_thompson5910

Question: which value of x makes the denominator, x-1, equal to 0?

7. anonymous

so confused

8. jim_thompson5910

if x-1 = 0, then what must x be?

9. anonymous

1

10. jim_thompson5910

since x = 1 makes x-1 = 0 true, that means x = 1 makes the denominator equal to zero. Agreed?

11. anonymous

yes

12. jim_thompson5910

to avoid dividing by zero, we have to kick out x = 1 from the domain

13. jim_thompson5910

now onto the next part

14. jim_thompson5910

when you solve $$\Large 2 - x \ge 0$$ for x, what do you get?

15. anonymous

i dont even know how to solve that, ive had sucky math teachers in high school they passed you based on if they liked you

16. jim_thompson5910

why not add x to both sides?

17. jim_thompson5910

|dw:1436580427007:dw|

18. jim_thompson5910

|dw:1436580444120:dw|

19. jim_thompson5910

what happens?

20. anonymous

2>0x

21. anonymous

well dont the x by the 0 just cancel cause its 0

22. jim_thompson5910

|dw:1436580524713:dw|

23. jim_thompson5910

|dw:1436580534366:dw| make sense?

24. anonymous

kinda

25. jim_thompson5910

so we know that $$\Large x \le 2$$ AND x cannot equal 1 as long as x is some number that satisfies those conditions above, then that number is allowed in the domain

26. anonymous

confused

27. jim_thompson5910

where?

28. anonymous

what you mean as long as x is some number that satisfies those conditions above, then that number is allowed in the domain

29. jim_thompson5910

if x is less than or equal to 2 AND x is not 1 then x is allowed in the domain

30. jim_thompson5910

those are the two conditions I'm talking about

31. anonymous

okay

32. anonymous

i just want to be done with this, ive had 3 hours of sleep, and been working since 6 this morning, and still have more to do ugh

33. jim_thompson5910

the problem is done at this point

34. anonymous

35. jim_thompson5910

what are the choices

36. anonymous

( , 1) (-1, 2] ( , 1) (1, 2] ( , 2) (1, 2] ( , -1) (1, 2] ( , -2) (1, 2] the first thing is a little infinity sign thing

37. jim_thompson5910

it would be $\Large (-\infty, 1) \cup (1, 2]$ if you wrote those conditions above in interval notation

38. jim_thompson5910

start with (-infinity, 2] and then kick out 1 from that interval

39. anonymous