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Calcmathlete

  • 2 years ago

Let f(x) be a quadratic function. If \(x^2 - 4x + 6 ≤ f(x) ≤ 2x^2 - 8x +10\) for all real numbers x, and f(12) = 182, find f(17).

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  1. TuringTest
    • 2 years ago
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    my first thought is to try factoring and see if that takes us somewhere

  2. Calcmathlete
    • 2 years ago
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    Aren't they both unfactorable?

  3. TuringTest
    • 2 years ago
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    I guess, I'm just brainstorming (and multitasking)

  4. TuringTest
    • 2 years ago
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    changing the second into 2(x^2-4x+5) may shed some light is thought #2

  5. TuringTest
    • 2 years ago
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    let's see\[f(12)=102\le182\le202\]

  6. TuringTest
    • 2 years ago
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    hm...

  7. TuringTest
    • 2 years ago
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    I think I need to stop multitasking to do this

  8. Calcmathlete
    • 2 years ago
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    \[227 ≤ f(17) ≤ 462\]

  9. TuringTest
    • 2 years ago
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    I have a strong feeling that @asnaseer is going to save us

  10. asnaseer
    • 2 years ago
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    could we make use of the fact that if the LHS=RHS then that fixes f(x)?

  11. asnaseer
    • 2 years ago
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    i.e.\[x^2-4x+6=2x^2-8x+10\]f we solve this we get:\[(x-2)^2=0\implies x=2\]

  12. asnaseer
    • 2 years ago
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    so at x=2 both the LHS (left hand side) and RHS (right hand side) are equal

  13. TuringTest
    • 2 years ago
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    interesting...

  14. Calcmathlete
    • 2 years ago
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    That would make f(2) = 2?

  15. Calcmathlete
    • 2 years ago
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    When you plug it back in anyway...

  16. asnaseer
    • 2 years ago
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    yes

  17. asnaseer
    • 2 years ago
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    and if we define f(x) as:\[f(x)=ax^2+bx+c\]then maybe we can form 3 equations and solve for a, b and c

  18. asnaseer
    • 2 years ago
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    we have two equations so far

  19. asnaseer
    • 2 years ago
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    maybe make use of f(0)?

  20. asnaseer
    • 2 years ago
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    which gives us:\[6\le c\le 10\]

  21. maheshmeghwal9
    • 2 years ago
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    \[x^2 - 4x + 6 ≤ (ax^2+bx+c)≤ 2x^2 - 8x +10\] \[\color{blue}{a(12)^2+b(12)+c}=182=\color{blue}{144a+12b+c.}\] \[\color{red}{a(17)^2+b(17)+c=?}\]\[6 \le c \le 10.\]

  22. Calcmathlete
    • 2 years ago
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    I just looked up how they did it...it's very strange, and I'm having a bit of trouble understanding a part of it...do you want me to post it or would you like to try figuring the rest out?

  23. asnaseer
    • 2 years ago
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    f(1) is also interesting, it gives us:\[3\le a+b+c\le4\]

  24. asnaseer
    • 2 years ago
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    what do you mean by /strange/?

  25. asnaseer
    • 2 years ago
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    is there a particular approach that is used?

  26. Calcmathlete
    • 2 years ago
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    Well, the way they did it is a lot simpler and way different from this way...they completed the square...

  27. asnaseer
    • 2 years ago
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    hmmm... let me think over that approach...

  28. maheshmeghwal9
    • 2 years ago
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    Becoz we all are doing this type of question first time. & they had done it already. I really appreciate @asnaseer 's work:D

  29. slaaibak
    • 2 years ago
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    all three equations have a turning point (minimum) at x=2. so you could use that.

  30. asnaseer
    • 2 years ago
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    I can't /see/ how completing the square helps. but I wonder if we can make use of my last result for f(1) to get 3 possible equations to solve.\[144a+12b+c=182\tag{1}\]\[4a+2b+c=2\tag{2}\]and then either:\[a+b+c=3\tag{3a}\]or:\[a+b+c=4\tag{3b}\]

  31. Spacelimbus
    • 2 years ago
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    Does it become a geometric question when they complete the square? |dw:1343584198825:dw|

  32. Calcmathlete
    • 2 years ago
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    They said that once you complete the square, you find that f(x) must have the same vertex as the other two equations and that you just have to solve for a.

  33. Spacelimbus
    • 2 years ago
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    just thinking out loud, thought they might subtract squares then.

  34. Spacelimbus
    • 2 years ago
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    hmm no, that's not what I thought then.

  35. Calcmathlete
    • 2 years ago
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    I don't unserstand how they MUST have the same vertex though...

  36. asnaseer
    • 2 years ago
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    interesting observation @slaaibak - maybe we can use that to get the 3rd equation?

  37. asnaseer
    • 2 years ago
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    yes @Calcmathlete that ties up with @slaaibak's observation

  38. slaaibak
    • 2 years ago
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    if we use f(x) = a(x-b)^2 + k, we can solve. because we have b:2 and k can be found by f(2) = k so it's pretty easy from here

  39. maheshmeghwal9
    • 2 years ago
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    ya so they all have same vertex. as @Calcmathlete 's.............. said:D

  40. Calcmathlete
    • 2 years ago
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    Thin is, I don't understand how they MUST all have the same vertex. Is it because of how the parabolas work when graphed?

  41. asnaseer
    • 2 years ago
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    the LHS and RHS have the same vertex which fixes the vertex of the middle one

  42. maheshmeghwal9
    • 2 years ago
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    no becoz they all have same minimum value at x=2

  43. maheshmeghwal9
    • 2 years ago
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    thus they all have same vertex

  44. slaaibak
    • 2 years ago
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    both have the same turning point. therefore, f(x) can't be decreasing if the other two functions are increasing, because f(2) are the same for all the equations. therefore all of them have a minimum at 2

  45. maheshmeghwal9
    • 2 years ago
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    |dw:1343584535907:dw|

  46. maheshmeghwal9
    • 2 years ago
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    3 parabolas

  47. maheshmeghwal9
    • 2 years ago
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    3 functions quadratic......

  48. asnaseer
    • 2 years ago
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    very similar to the fact that x=2 gives the same value for LHS and RHS which fixes the value of f(2)

  49. Calcmathlete
    • 2 years ago
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    Alright. That's really why i didn't understand. I guess since the opposite ends of the compound inequality have the same vertex, it is justified. From there it is easy I guess. \[f(12) = a(x - 2)^2 + 2 \implies a = 1.8\]Then we can just plug in 17? \[f(17) = 1.8(225) + 2 = 405 Is that right?

  50. slaaibak
    • 2 years ago
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    f(2) = 2 f(x) = a(x-2)^2 + k b=2 due to minimum so f(2) = a(2-2)^2 + k = 2 therefore k = 2 now f(12) = 182 so 182 = a(12-2)^2 + 2 182 = 100a + 2 100a = 180 a = 1.8

  51. asnaseer
    • 2 years ago
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    yes @Calcmathlete

  52. slaaibak
    • 2 years ago
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    yeah that's correct

  53. Calcmathlete
    • 2 years ago
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    Alright. Thank you! I wish I could give multiple medals...

  54. asnaseer
    • 2 years ago
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    yw :) and yes, I wish we had Olympic-style medals - Gold/Silver/Bronze :)

  55. asnaseer
    • 2 years ago
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    but then we would have to drug test everyone to make sure they were not cheating! :D

  56. Calcmathlete
    • 2 years ago
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    lol...

  57. maheshmeghwal9
    • 2 years ago
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    Graphs must be there.

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