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anonymous

  • one year ago

explain how to find the maximum value for each function and determine which function has the largest maximum y-value.

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    @UnkleRhaukus

  3. UnkleRhaukus
    • one year ago
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    The function is \[g(x) = 2\cos(2x-\pi)+4\] This will have maximum values when the cosine term is at maximums

  4. anonymous
    • one year ago
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    f(x) = −4(x − 6)^2 + 3

  5. UnkleRhaukus
    • one year ago
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    do you know what the largest value of \(\cos\theta\), can ever be? (for all possible angles \(\theta\) )

  6. anonymous
    • one year ago
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    no

  7. UnkleRhaukus
    • one year ago
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    remember that cosine is the ratio of the adjacent side in a right angled triangle, to its hypotenuse, (note that the hypotenuse is always the largest side)

  8. anonymous
    • one year ago
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    ok ok , i get that

  9. UnkleRhaukus
    • one year ago
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    |dw:1440303568385:dw|

  10. UnkleRhaukus
    • one year ago
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    The cosine function (like the sine function ) oscillates between its maximum and minimum values ; ±1

  11. anonymous
    • one year ago
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    okk

  12. anonymous
    • one year ago
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    So the maximun for g(x) would be 1?

  13. UnkleRhaukus
    • one year ago
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    not quite, that max for cos theta, and hence the max for cos(2x-π) is 1

  14. UnkleRhaukus
    • one year ago
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    so the max for g(x) = 2cos(2x-π) + 4 is gmax = 2(+1) + 4 =

  15. anonymous
    • one year ago
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    so 7 lol

  16. UnkleRhaukus
    • one year ago
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    (whereas the minimum is gmin = 2(-1) + 4 = -2+4 = 2 )

  17. UnkleRhaukus
    • one year ago
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    wait gmax = 2(+1) + 4 = 2 times 1 + 4 =

  18. anonymous
    • one year ago
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    oh i was thinking adding. silly me lol. so 6

  19. UnkleRhaukus
    • one year ago
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    notice that we can see in the plot, that the blue line is oscillating between 2 and 6, which agrees

  20. UnkleRhaukus
    • one year ago
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    so now lets look at \[f(x) = −4(x − 6)^2 + 3 \]

  21. anonymous
    • one year ago
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    how would we set that up

  22. UnkleRhaukus
    • one year ago
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    What can you tell me about f(x)?

  23. UnkleRhaukus
    • one year ago
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    (what is the parent function of f )

  24. anonymous
    • one year ago
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    i dont get it?

  25. UnkleRhaukus
    • one year ago
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    its a second order polynomial, right? so its a quadratic equation, and will look like some sort of parabola

  26. anonymous
    • one year ago
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    is this calculus? if so, take the derivative of the function, and set to 0 to find where it has a max (or min). However, you do not need calculus. The first function, f(x)=-4(x-6)^2 + 3 is a parabola ( in the shape of a "frown" ) with a max value at its vertex. because they gave you the equation in "vertex form" y = a(x-h)^2 + k , you can read off the vertex to be (h,k) In this case, the vertex is at (6,3) for the second function, g(x)=2cos(2x-pi)+4 you should know that the max value of the cosine is 1, so the max will be g(x)= 2*1 + 4 = 6

  27. UnkleRhaukus
    • one year ago
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    @TavTav are you still here?

  28. anonymous
    • one year ago
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    yes sorry

  29. anonymous
    • one year ago
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    so i got g(x) but how would you get f(x)

  30. UnkleRhaukus
    • one year ago
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    given that f(x) is some sort of parabola, do you expect f to have both max and min values?

  31. anonymous
    • one year ago
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    no

  32. UnkleRhaukus
    • one year ago
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    \[f(x)=-4(x-6)^2 + 3\] which is a bit like: \(-x^2\) which do you expect: a max, or a min value?

  33. anonymous
    • one year ago
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    max

  34. UnkleRhaukus
    • one year ago
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    good, and the max will be when the terms under the square, will be what value?

  35. anonymous
    • one year ago
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    the max value would be the 3?

  36. UnkleRhaukus
    • one year ago
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    yeah, the max is when the terms under the square is zero, \[f_\text{max}=-4(0)^2 + 3\\ \qquad=-0+3\\ \qquad=3\]

  37. anonymous
    • one year ago
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    thank you. i need to have it in complete sentences so can you help me form those?

  38. anonymous
    • one year ago
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    for both g(x) and f(x)

  39. UnkleRhaukus
    • one year ago
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    show me what you got

  40. anonymous
    • one year ago
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    uum ok

  41. anonymous
    • one year ago
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    lol

  42. UnkleRhaukus
    • one year ago
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    The general idea was that we considered the parent functions \(G(X) = \cos (X)\), and \(F(X)=-X^2\), respectively

  43. anonymous
    • one year ago
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    I know how we got g(x) i just dont know how to put it in words for f(x) we used the zeros of the equation to find the maximun value?

  44. UnkleRhaukus
    • one year ago
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    we didn't really use the "zeroes of the equations", they are different things

  45. anonymous
    • one year ago
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    ohh , uhh. so how would you word it

  46. anonymous
    • one year ago
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    that we compared the functions to their parent functions to get the maximum value

  47. UnkleRhaukus
    • one year ago
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    yeah, something like that, we know how to find the max/min of the parent functions, and then we apply these results to our particular functions

  48. UnkleRhaukus
    • one year ago
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    scaling and shifting as specified

  49. anonymous
    • one year ago
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    so i can use that as the answer?

  50. UnkleRhaukus
    • one year ago
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    kind of. you might like to write separate paragraphs for each function

  51. anonymous
    • one year ago
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    okok , thank you so much!! you made this easy to understand.your a lifesaver lol

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