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anonymous

  • one year ago

Will METAL!! Find the domain of the given function. f(x) = square root of quantity x plus three divided by quantity x plus eight times quantity x minus two. x > 0 All real numbers x ≥ -3, x ≠ 2 x ≠ -8, x ≠ -3, x ≠ 2

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  1. Directrix
    • one year ago
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    Would you write this out in symbols? I am not following what this equation is: f(x) = square root of quantity x plus three divided by quantity x plus eight times quantity x minus two.

  2. phi
    • one year ago
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    Do you see the equation editor in the bottom left? that would be good to use

  3. anonymous
    • one year ago
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    \[\sqrt{x+3}/(x+8)(x-2)\]

  4. anonymous
    • one year ago
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    f(x)= ^

  5. phi
    • one year ago
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    meanwhile, the domain are all "legal" x values. In other words, all x values that do not cause trouble. What kinds of trouble? we don't want square roots of negative numbers. if an x value causes that, we don't allow x to be that value no divide by 0: if an x value causes us to divide by 0, we don't allow that. Use those rules to decide which x values are not allowed, and that tells you which are allowed.

  6. phi
    • one year ago
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    to answer, first look at \( \sqrt{x+3}\) any idea what x values might cause us to have the square root of a negative number ?

  7. anonymous
    • one year ago
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    no not sure.

  8. phi
    • one year ago
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    negative means less than zero if x+3 < 0 then that means x+3 is negative. And we don't want that we want x+3 >=0 (that means x+3 is 0 or positive) add -3 to both sides , what do you get ?

  9. anonymous
    • one year ago
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    if you add -3 to both sides then you get x>-3?

  10. phi
    • one year ago
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    you mean x>=-3 in other words, if we want x+3 to be zero or bigger we need x+3 >= 0 and that means x >= -3 (otherwise we would have a square root of a negative number, and we don't want that)

  11. phi
    • one year ago
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    so \( x\ge -3 \) is part of the answer. next, look that the "bottom" (x+8)(x-2) what happens if x is -8 ?

  12. anonymous
    • one year ago
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    oh I see it, wow your good.

  13. phi
    • one year ago
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    the bottom becomes (-8+8)(-8-2) and -8+8 is 0 0 times anything is 0 in other words the bottom becomes 0, and we would be dividing the top by 0 we don't allow divide by zero. so we can't let x be -8 luckily , x must be >= -3 (to keep the top legal), so we are ok (-8 is smaller than -3, so we won't allow it) but we have (x+8)(x-2) what happens if x= 2 ?

  14. anonymous
    • one year ago
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    (2+8)(2-2)

  15. anonymous
    • one year ago
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    2-2=0

  16. anonymous
    • one year ago
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    0 times anything = 0

  17. phi
    • one year ago
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    yes, and we have zero. we don't want zero in the bottom so we have to say x is not allowed to be 2 all together, the allowed x values are \[ x\ge-3 , x\ne 2\] which means x is -3 or bigger, but not 2

  18. anonymous
    • one year ago
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    can we do another?

  19. phi
    • one year ago
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    ok

  20. anonymous
    • one year ago
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    Verify the identity. cos 4x + cos 2x = 2 - 2 sin^2 2x - 2 sin^2 x

  21. phi
    • one year ago
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    It looks painful. Do you have the "half-angle" formula for cos 4x ?

  22. anonymous
    • one year ago
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    isn't it =+-square root 1+ cos theta/ 2

  23. phi
    • one year ago
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    there are 3 different versions \[ \cos(2a) = \cos^2(a) - \sin^2(a) = 2\cos^2(a)-1 = 1 - 2 \sin^2(a) \]

  24. phi
    • one year ago
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    because in your problem you only have sin on the right side, I would use the last version \[ \cos(2a) = 1 - 2 \sin^2(a) \] can you use that to replace cos(4x) in your problem ?

  25. phi
    • one year ago
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    match up cos 4x with cos 2a write cos 4x as cos (2* 2x) and we match cos (2 * a) a is 2x in your problem

  26. phi
    • one year ago
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    \[ \cos(2a) = 1 - 2 \sin^2(a) \\ \cos(4x) = ? \]

  27. anonymous
    • one year ago
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    2-2sin^2(a) ?

  28. phi
    • one year ago
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    \[ \cos(2a) = 1 - 2 \sin^2(a) \\ \cos(4x) = 1 - 2 \sin^2(2x) \]

  29. anonymous
    • one year ago
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    so 1-2sin^2(2x) 1-2sin^2(2x) + cos 2x = 2 - 2 sin^2 2x - 2 sin^2 x

  30. phi
    • one year ago
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    yes, and if we add +2sin^2(2x) to both sides, we can simplify that a bit

  31. anonymous
    • one year ago
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    1-2sin^2(2x) + cos 2x = 2 - 2 sin^2 2x - 2 sin^2 x +2sin^2x +2sin^2x ---------------------------------------------- ?

  32. phi
    • one year ago
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    yes

  33. anonymous
    • one year ago
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    1-2sin^2(2x) + cos 2x = 2 - 2 sin^2 2x - 2 sin^2 x +2sin^2x +2sin^2x ---------------------------------------------- 1(2x) + cos 2x = 2-2 sin^2 (2x) I'm not sure

  34. phi
    • one year ago
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    oops, you should add 2 sin^2(2x) (not 2sin^2(x) )

  35. phi
    • one year ago
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    1-2sin^2(2x) + cos 2x = 2 - 2 sin^2(2x) - 2 sin^2 x +2sin^2(2x) +2 sin^2(2x) ---------------------------------------------- 1 + cos(2x) = 2 - 2 sin^2(x)

  36. phi
    • one year ago
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    It looks like we should use the same rule as before , but now on cos(2x)

  37. anonymous
    • one year ago
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    1-2sin^2(x)

  38. phi
    • one year ago
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    yes, put that in for cos(2x)

  39. anonymous
    • one year ago
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    1 + cos(2x) = 2 - 2 sin^2(x) ------------------------------------------------ 1-2sin^2(x)= -2sin^2(x)

  40. phi
    • one year ago
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    you replace cos(2x) with 1-2sin^2(x) like this 1 + cos(2x) = 2 - 2 sin^2(x) 1 + (1-2sin^2(2x)) = 2 - 2 sin^2(2x) now simplify the left side

  41. phi
    • one year ago
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    the left side is \[ 1 + 1 - 2\sin^2(2x) \]

  42. anonymous
    • one year ago
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    1 + (1-2sin^2(2x)) = 2 - 2 sin^2(2x) ------------------------------- 1+1-2sin^2(2x) = 2-2sin^2(2x) -------------------------------- 2-2sin^2(2x) = 2-2sin^2(2x) yay!

  43. phi
    • one year ago
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    yes, both sides are obviously the same.

  44. anonymous
    • one year ago
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    can we do another problem, not like this one though.

  45. phi
    • one year ago
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    one more, but please make it a new post.

  46. anonymous
    • one year ago
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    ok hold on.

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