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whadduptori

Can somebody just check me? For the function f(x) = (3 – 4x)², find f–1. Determine whether f–1 is a function. Is this right? what i have: f(x)=(3-4x)^2 f(1)=(3-4x1)^2 =(3-4)^2 =(-1)^2 =1 f is a function

  • one year ago
  • one year ago

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  1. whpalmer4
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    Aren't you supposed to find the inverse?

    • one year ago
  2. whadduptori
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    im just so confuse.... i have no idea what im doing

    • one year ago
  3. whpalmer4
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    Is it written like this? \[f(x) = (3-4x)^2, find f^{-1}(x)\]

    • one year ago
  4. whadduptori
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    yes!

    • one year ago
  5. whpalmer4
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    Okay, yeah, that's finding the inverse function and checking if that is a function, not evaluating the original function at f(1) or f(-1)... So, do you know how to find the inverse function?

    • one year ago
  6. whadduptori
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    ohh1! nope i honestly do not remember

    • one year ago
  7. whadduptori
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    wait you switch the x and y right?

    • one year ago
  8. whpalmer4
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    Okay, if you want to find the inverse function, you write out the original function as y = f(x) = (3-4x)^2, but then you swap x and y, and rearrange it as y = <some function> For example, if you had y = f(x) = 3x, f^(-1)(x) would be found like this: y = 3x x = 3y x/3 = y so y = x/3 would be the inverse of y = 3x.

    • one year ago
  9. whadduptori
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    ok so in my example would it be.. y = (3 – 4x)² x = (3-4y)^2

    • one year ago
  10. whpalmer4
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    so after the dust settles, what does y = ?

    • one year ago
  11. whadduptori
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    so x = (3-4y)^2 y= 1/4 (3-sqrt(x)) so... f^-1 = 1/4 (3-sqrt(x))

    • one year ago
  12. whpalmer4
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    that's half of it...

    • one year ago
  13. whpalmer4
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    Won't 1/4 (3 + sqrt(x)) = y also?

    • one year ago
  14. whadduptori
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    yes so it would be y = (1/4)( 3 ± sqrt(x))

    • one year ago
  15. whpalmer4
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    Right. So, is the inverse a function?

    • one year ago
  16. whadduptori
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    f-1(x) = y = (1/4)(3 ± sqrt(x)) ??

    • one year ago
  17. whadduptori
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    oh no sorry its not a function!

    • one year ago
  18. whpalmer4
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    That's right, it fails the vertical line test. A vertical line passes through the curve more than once, so there is not a unique value of y for every x.

    • one year ago
  19. whpalmer4
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    If you draw a line through the origin heading up and to the right at a 45 degree angle (slope = 1), the inverse function is simply the original function reflected along that line. So if you have a parabola opening upward centered on the y-axis, the inverse is a parabola opening to the right centered on the x-axis.

    • one year ago
  20. whpalmer4
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    • one year ago
  21. whpalmer4
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    And a little thought about that suggests that functions more complicated than straight lines may not have an inverse which is a function...

    • one year ago
  22. whadduptori
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    oh okay! i see what your saying now! you explained it to me well! thank you for your help!! (:

    • one year ago
  23. whpalmer4
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    Glad I could help make the little bulb come to life :-)

    • one year ago
  24. whpalmer4
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    little light bulb that is

    • one year ago
  25. whadduptori
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    hahah you are awesome! can you maybe help me with one more?

    • one year ago
  26. whpalmer4
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    I can try!

    • one year ago
  27. whadduptori
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    Use the Change of Base Formula to evaluate log7(76)

    • one year ago
  28. whadduptori
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    like would it be log7(76)= log(76)/log10(7) is that the answer?

    • one year ago
  29. whpalmer4
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    \[\log_b a = \frac{\log a }{\log b}\] So \[\log_7 76 = \frac{\log{76}}{\log 7}\]

    • one year ago
  30. whpalmer4
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    Both of the logs on the right hand side need to be in the same base, of course. Do you know how you can show that this is true?

    • one year ago
  31. whadduptori
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    wait so would it be log76 = 1.88081359 and log 7= 84509804 and you would divide those right?

    • one year ago
  32. whpalmer4
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    Yes, but I think you are missing a decimal in your value of log 7...

    • one year ago
  33. whadduptori
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    oh yeah i did! so it would .84509804

    • one year ago
  34. whadduptori
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    so the answer would be 2.22556685

    • one year ago
  35. whpalmer4
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    \[\log_b a = \frac{\log a}{\log b}\]Raise both sides as powers of b \[b^{\log_b a} = b^{\frac{\log a}{\log b}}\]But the left half of that is simply a by definition \[a = b^{\frac{\log a}{\log b}}\]and if we take the log of both sides \[\log a = \frac{\log a}{\log b}*\log b\]because \[\log u^n = n \log u\]

    • one year ago
  36. whpalmer4
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    As a quick check of your answer, 7^2*(7*1/5) = 68.6, and 7^2*(7*1/4) = 85 so it looks like the answer is in the right ballpark. 7^2.225 is a little bit bigger than 1/5, and a little smaller than 1/4.

    • one year ago
  37. whadduptori
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    so my answer was right?

    • one year ago
  38. whpalmer4
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    Yes!

    • one year ago
  39. whadduptori
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    ah thank you again!!!!!

    • one year ago
  40. whpalmer4
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    Now you can go tutor all of your friends :-) That's a good way to reinforce your grasp of the material — try to explain it to someone else. Works best if they can ask questions in return, so explaining it to the cat or dog might not be as effective :-)

    • one year ago
  41. whadduptori
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    I think I will! you are amazing thank you again, and I usually get confused but you made me understand!! :)

    • one year ago
  42. whadduptori
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    \[\sqrt{7}(\sqrt{x}-7\sqrt{7}\]

    • one year ago
  43. whpalmer4
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    Is that supposed to be \[\sqrt{7}(\sqrt{x}-7\sqrt{7})\] and is the goal to simplify it?

    • one year ago
  44. whadduptori
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    oh yes sorry! Multiply and simplify if possible.

    • one year ago
  45. whpalmer4
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    Well, if you distribute it out, what do you get?

    • one year ago
  46. whadduptori
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    \[\sqrt{7x}-49\sqrt{x}\]

    • one year ago
  47. whpalmer4
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    where did the \sqrt{x} next to the 49 come from?

    • one year ago
  48. whadduptori
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    i have no idea.. how do you distribute that?

    • one year ago
  49. whpalmer4
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    Well, multiply sqrt(7) by the first term in the parentheses, and you get sqrt(7)sqrt(x). Multiply sqrt(7) by the second term in the parentheses, and you get -sqrt(7)*7*sqrt(7), right?

    • one year ago
  50. whpalmer4
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    and that last bit reduces to -49

    • one year ago
  51. whpalmer4
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    \[\sqrt{7x}-49 = \sqrt{7}\sqrt{x}-49\] (I'd probably write the first one if it was a final answer, the second if I was going to do further canceling or manipulation).

    • one year ago
  52. whpalmer4
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    \[a(b-c) = ab - ac, a = \sqrt{7}, b=\sqrt{x}, c = 7\sqrt{7}\]

    • one year ago
  53. whadduptori
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    wait whats the answer.. im lost

    • one year ago
  54. whpalmer4
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    See my post with the (I'd probably write ...) The last bit was just showing why the second term shouldn't have an x in it

    • one year ago
  55. whadduptori
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    Oh I see what you were doing now! sorry

    • one year ago
  56. whpalmer4
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    Not a problem. If we were talking, I would have said a bit more, and you would have understood right away.

    • one year ago
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