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livelaughlilz

  • 3 years ago

what is the minimum value of 7^(x^2-4x+7)

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  1. saifoo.khan
    • 3 years ago
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    \[\Large 7^{x^2-4x+7}\]Like this?

  2. livelaughlilz
    • 3 years ago
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    yes

  3. saifoo.khan
    • 3 years ago
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    @mathmate @UnkleRhaukus

  4. saifoo.khan
    • 3 years ago
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    Minimum value should be there where dy/dx = 0.

  5. livelaughlilz
    • 3 years ago
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    what do you mean by the "d"?

  6. saifoo.khan
    • 3 years ago
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    In other words minimum value occurs where the derivative of the function is equal to zero.

  7. livelaughlilz
    • 3 years ago
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    isn't derivatives calculus? i haven't learned that yet and i don't think we're expected to know it to solve this problem

  8. saifoo.khan
    • 3 years ago
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    http://www.mememaker.net/static/images/templates/14288.jpg

  9. livelaughlilz
    • 3 years ago
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    haha i was thinking about it and wouldn't it be one over infinity? because if x^2-4x+7 was negative infinity than 7^(x^2-4x+7) would be 1/infinity

  10. mathmate
    • 3 years ago
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    Hint: 7^x is an increasing function. So if we minimize x, 7^x is also a minimum.

  11. mathmate
    • 3 years ago
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    x^2-4x+7 cannot be at -inf. It is a parabola concave upwards.!

  12. saifoo.khan
    • 3 years ago
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    And are we going to try that by trial and error? @mathmate

  13. KingGeorge
    • 3 years ago
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    No, we need to simply find the vertex of the parabola to find the minimum value of the parabola.

  14. saifoo.khan
    • 3 years ago
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    Howw??

  15. mathmate
    • 3 years ago
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    Good point!] @livelaughlilz can you take it from here?

  16. mathmate
    • 3 years ago
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    Completing the squares will do the job.

  17. livelaughlilz
    • 3 years ago
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    yeah the vertex of the parabola is (2,3)

  18. livelaughlilz
    • 3 years ago
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    (x-2)^2=y-3

  19. mathmate
    • 3 years ago
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    x^2-4x+7=(x-2)^2+3 so the vertex is at (2,3)

  20. saifoo.khan
    • 3 years ago
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    http://i1.kym-cdn.com/entries/icons/original/000/009/993/tumblr_m0wb2xz9Yh1r08e3p.jpg

  21. mathmate
    • 3 years ago
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    Good job livelaughlilz! Speed+accuracy!

  22. mathmate
    • 3 years ago
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    I like his movies! :)

  23. livelaughlilz
    • 3 years ago
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    oh so the answer is 343. the lowest value of x is 2, and if you plug it in you get 7^3=343

  24. saifoo.khan
    • 3 years ago
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    Yes^ @livelaughlilz

  25. mathmate
    • 3 years ago
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    7^2 = ???

  26. saifoo.khan
    • 3 years ago
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    @mathmate : haha, nice. My brother just love his movies. i don't. :/

  27. mathmate
    • 3 years ago
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    Nevermind, I was out of my mind. Yes, 7^3=343

  28. saifoo.khan
    • 3 years ago
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    You were thinking about Jackie Chan. :D

  29. KingGeorge
    • 3 years ago
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    As a sidenote @mathmate, you can find the x-coordinate of the vertex using the formula \[v_x=\frac{-b}{2a}\]

  30. mathmate
    • 3 years ago
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    Thank you for the defence, even though it was a lame one! Good thinking! :)

  31. saifoo.khan
    • 3 years ago
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    I simply love this method. Satellite73 taught me this. -b/2a

  32. livelaughlilz
    • 3 years ago
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    to add onto @KingGeorge the y-coordinate of the vertex is -[(b^2-4ac)/(4a)]

  33. mathmate
    • 3 years ago
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    Thank you @KingGeorge, my memory is limited and sometimes defective. But thanks for the shortcut anyway!

  34. livelaughlilz
    • 3 years ago
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    you can figure all that out by completing the square of y=ax^2+bx+c

  35. livelaughlilz
    • 3 years ago
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    and making y=ax^2+bx+c into vertex form

  36. mathmate
    • 3 years ago
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    Actually I know the -b/2a part, that's how I get to do the completing the square. It's just I never managed to memorize the -[(b^2-4ac)/(4a)] part.

  37. livelaughlilz
    • 3 years ago
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    well b^2-4ac is also the discriminant...

  38. saifoo.khan
    • 3 years ago
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    @livelaughlilz is a boss at this.

  39. mathmate
    • 3 years ago
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    Perhaps from this time on I will remember. Practice makes perfect! :)

  40. mathmate
    • 3 years ago
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    Actually what I do is finding it by c-(b/2a)^2, which is exactly what we do when we complete the square.

  41. livelaughlilz
    • 3 years ago
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    oh

  42. livelaughlilz
    • 3 years ago
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    i'm going to close this question now. thanks everyone!

  43. mathmate
    • 3 years ago
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    See you all!

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