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whadduptori

  • 2 years ago

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

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  1. whpalmer4
    • 2 years ago
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    Aren't you supposed to find the inverse?

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

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

  4. whadduptori
    • 2 years ago
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    yes!

  5. whpalmer4
    • 2 years ago
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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?

  6. whadduptori
    • 2 years ago
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    ohh1! nope i honestly do not remember

  7. whadduptori
    • 2 years ago
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    wait you switch the x and y right?

  8. whpalmer4
    • 2 years ago
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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.

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

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

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

  12. whpalmer4
    • 2 years ago
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    that's half of it...

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

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

  15. whpalmer4
    • 2 years ago
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    Right. So, is the inverse a function?

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

  17. whadduptori
    • 2 years ago
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    oh no sorry its not a function!

  18. whpalmer4
    • 2 years ago
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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.

  19. whpalmer4
    • 2 years ago
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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.

  20. whpalmer4
    • 2 years ago
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  21. whpalmer4
    • 2 years ago
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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...

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

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

  24. whpalmer4
    • 2 years ago
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    little light bulb that is

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

  26. whpalmer4
    • 2 years ago
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    I can try!

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

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

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

  30. whpalmer4
    • 2 years ago
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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?

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

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

  33. whadduptori
    • 2 years ago
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    oh yeah i did! so it would .84509804

  34. whadduptori
    • 2 years ago
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    so the answer would be 2.22556685

  35. whpalmer4
    • 2 years ago
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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 \[\log u^n = n \log u\]

  36. whpalmer4
    • 2 years ago
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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.

  37. whadduptori
    • 2 years ago
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    so my answer was right?

  38. whpalmer4
    • 2 years ago
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    Yes!

  39. whadduptori
    • 2 years ago
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    ah thank you again!!!!!

  40. whpalmer4
    • 2 years ago
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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 :-)

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

  42. whadduptori
    • 2 years ago
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    \[\sqrt{7}(\sqrt{x}-7\sqrt{7}\]

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

  44. whadduptori
    • 2 years ago
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    oh yes sorry! Multiply and simplify if possible.

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

  46. whadduptori
    • 2 years ago
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    \[\sqrt{7x}-49\sqrt{x}\]

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

  48. whadduptori
    • 2 years ago
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    i have no idea.. how do you distribute that?

  49. whpalmer4
    • 2 years ago
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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?

  50. whpalmer4
    • 2 years ago
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    and that last bit reduces to -49

  51. whpalmer4
    • 2 years ago
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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).

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

  53. whadduptori
    • 2 years ago
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    wait whats the answer.. im lost

  54. whpalmer4
    • 2 years ago
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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

  55. whadduptori
    • 2 years ago
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    Oh I see what you were doing now! sorry

  56. whpalmer4
    • 2 years ago
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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.

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