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anonymous

  • one year ago

2. Let f be the function shown below, with domain the closed interval [0, 6]. The graph of f is shown below, composed of two semicircles. Let h(x) = ∫ f (t)dt from x=0 to x=(2x-1). ( I don't think the image will load... it is a semicircle above the x-axis with r=2 and center (2,0) connected to a semicircle below the x-axis with r=1 and center (5,0).) A. Determine the domain of h(x). For this, I solved 1/2<=x<=7/2 B. Find h′(5/2) ... this is where I am lost, I don't know how to find the derivative of this unknown function. C. At what x is h(x) a maximum? Show all the analysis tha

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  1. Michele_Laino
    • one year ago
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    is your function like this: |dw:1435084959591:dw|

  2. anonymous
    • one year ago
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    Yes

  3. Michele_Laino
    • one year ago
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    we can write this: \[\Large f\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {\sqrt {4 - {{\left( {t - 2} \right)}^2}} ,\quad 0 \leqslant t \leqslant 4} \\ { - \sqrt {1 - {{\left( {t - 5} \right)}^2}} ,\quad 4 \leqslant t \leqslant 5} \end{array}} \right.\]

  4. Michele_Laino
    • one year ago
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    so we have: \[\Large h\left( x \right) = \int_0^{2x - 1} {f\left( t \right)\;dt} \]

  5. Michele_Laino
    • one year ago
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    now, we have to distinguish these cases: \[0 \leqslant 2x - 1 < 4\]

  6. Michele_Laino
    • one year ago
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    and: \[4 \leqslant 2x - 1 \leqslant 6\]

  7. Michele_Laino
    • one year ago
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    first case: \[1 \leqslant x < \frac{5}{2}\]

  8. Michele_Laino
    • one year ago
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    we have: \[\large h\left( x \right) = \int_0^{2x - 1} {f\left( t \right)\;dt} = \int_0^{2x - 1} {\sqrt {4 - {{\left( {t - 2} \right)}^2}} \;dt} \]

  9. Michele_Laino
    • one year ago
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    second case: \[\frac{5}{2} \leqslant x \leqslant \frac{7}{2}\]

  10. Michele_Laino
    • one year ago
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    here we can write: \[\large h\left( x \right) = \int_0^{2x - 1} {f\left( t \right)\;dt} = - \int_0^{2x - 1} {\sqrt {1 - {{\left( {t - 5} \right)}^2}} \;dt} \]

  11. Michele_Laino
    • one year ago
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    both integral can be solved using a trigonometric substitution, namely, we have to substitute \[t - 2 = 2\sin \theta \] for first integral, and \[t - 5 = \sin \theta \] for the second one

  12. Michele_Laino
    • one year ago
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    integrals*

  13. Michele_Laino
    • one year ago
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    I'm computing the first integral...

  14. Michele_Laino
    • one year ago
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    please wait...

  15. Michele_Laino
    • one year ago
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    for first integral I got this expression: \[2\left\{ {\arcsin \left( {\frac{{2x - 3}}{2}} \right) + \frac{{2x - 3}}{2}\sqrt {1 - {{\left( {\frac{{2x - 3}}{2}} \right)}^2}} } \right\}\]

  16. anonymous
    • one year ago
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    But we want the second integral to solve for h'(5/2), yes? Am I right about that?

  17. Michele_Laino
    • one year ago
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    yes! Nevertheless we have to determine the domain of both integrals

  18. Michele_Laino
    • one year ago
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    now the first integral can exist if and only if: \[ - 1 \leqslant \frac{{2x - 3}}{2} \leqslant 1\]

  19. Michele_Laino
    • one year ago
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    from which we get: \[\frac{1}{2} \leqslant x \leqslant \frac{5}{2}\]

  20. anonymous
    • one year ago
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    so only between x=1/2 and x=2.5? So is that its domain?

  21. Michele_Laino
    • one year ago
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    now let's go to compute the second integral

  22. Michele_Laino
    • one year ago
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    The first integral is computed neglecting a constant term

  23. Michele_Laino
    • one year ago
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    for second integral I got this: \[\begin{gathered} h\left( x \right) = \int_5^{2x - 1} {f\left( t \right)\;dt} = - \int_5^{2x - 1} {\sqrt {1 - {{\left( {t - 5} \right)}^2}} \;dt} = \hfill \\ \hfill \\ = \frac{1}{2}\left\{ {\arcsin \left( {2x - 6} \right) + \left( {2x - 6} \right)\sqrt {1 - {{\left( {2x - 6} \right)}^2}} } \right\} \hfill \\ \hfill \\ \end{gathered} \]

  24. Michele_Laino
    • one year ago
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    your answer is right, the interval (0.5, 2.5) is part of the domain of h(x)

  25. Michele_Laino
    • one year ago
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    now the second integral exists if and only if: \[ - 1 \leqslant 2x - 6 \leqslant 1\]

  26. anonymous
    • one year ago
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    But that's only for the part associated with the first integral, correct? The whole domain is 1/2<=x<=7/2

  27. Michele_Laino
    • one year ago
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    or equivalently: \[\frac{5}{2} \leqslant x \leqslant \frac{7}{2}\]

  28. Michele_Laino
    • one year ago
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    that's right, the domain of h(x) is given by the union of both sets, namely: \[\frac{1}{2} \leqslant x \leqslant \frac{7}{2}\]

  29. Michele_Laino
    • one year ago
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    now we have to determine h'(5/2)

  30. Michele_Laino
    • one year ago
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    using the fundamental theorem of integral calculus, we can write this:

  31. Michele_Laino
    • one year ago
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    we have to compute directly the first derivative of both expressions above

  32. Michele_Laino
    • one year ago
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    I'm computing, please wait...

  33. Preetha
    • one year ago
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    Michele, what a great explanation. TSV, please be sure to rate when done, so Michele can get their share of OwlBucks

  34. Michele_Laino
    • one year ago
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    I got this: \[\Large h'\left( x \right) = \frac{4}{{\sqrt {1 - {{\left( {\frac{{2x - 3}}{2}} \right)}^2}} }},\quad \frac{1}{2} \leqslant x < \frac{5}{2}\quad \]

  35. Michele_Laino
    • one year ago
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    Thanks! Mrs. Preetha

  36. Michele_Laino
    • one year ago
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    now I compute h'(x) for the second expression:

  37. anonymous
    • one year ago
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    @preetha I definitely will! I'm very grateful for the help on this problem, I'm doing some calc review and struggled with this concept! So this is very helpful for me!

  38. Michele_Laino
    • one year ago
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    please wait, I'm computing...

  39. Michele_Laino
    • one year ago
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    \[h'\left( x \right) = 2\sqrt {1 - {{\left( {2x - 6} \right)}^2}} ,\quad \frac{5}{2} \leqslant x \leqslant \frac{7}{2}\]

  40. Michele_Laino
    • one year ago
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    sorry the first equation for h'(x) is wrong, here is the right equation: \[h'\left( x \right) = 4\sqrt {1 - {{\left( {\frac{{2x - 3}}{2}} \right)}^2}} ,\quad \frac{1}{2} \leqslant x < \frac{5}{2}\quad \]

  41. Michele_Laino
    • one year ago
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    now, in order to find h'(5/2), we have to evaluate both expression above at x=5/2. So if we substitute x=5/2, we get: \[h'\left( {\frac{5}{2}} \right) = 4\sqrt {1 - {{\left( {\frac{{2 \times \frac{5}{2} - 3}}{2}} \right)}^2}} = 0\]

  42. Michele_Laino
    • one year ago
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    whereas the second expression, gives: \[h'\left( {\frac{5}{2}} \right) = 2\sqrt {1 - {{\left( {2 \times \frac{5}{2} - 6} \right)}^2}} = 0\]

  43. Michele_Laino
    • one year ago
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    so we can write: \[h'\left( {\frac{5}{2}} \right) = 0,\] in the ordinary sense of the first derivative

  44. Michele_Laino
    • one year ago
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    now let's go to the third part of your question

  45. Michele_Laino
    • one year ago
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    here we have to solve these equations: \[h'\left( x \right) = 4\sqrt {1 - {{\left( {\frac{{2x - 3}}{2}} \right)}^2}} = 0,\quad \frac{1}{2} \leqslant x < \frac{5}{2}\quad \]

  46. Michele_Laino
    • one year ago
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    and \[h'\left( x \right) = 2\sqrt {1 - {{\left( {2x - 6} \right)}^2}} = 0,\quad \frac{5}{2} \leqslant x \leqslant \frac{7}{2}\]

  47. Michele_Laino
    • one year ago
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    oops.. the second equation is: \[\begin{gathered} h'\left( x \right) = - 2\sqrt {1 - {{\left( {2x - 6} \right)}^2}} = 0,\quad \frac{5}{2} \leqslant x \leqslant \frac{7}{2} \hfill \\ \hfill \\ \end{gathered} \]

  48. Michele_Laino
    • one year ago
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    since we have: \[\begin{gathered} h\left( x \right) = \int_5^{2x - 1} {f\left( t \right)\;dt} = - \int_5^{2x - 1} {\sqrt {1 - {{\left( {t - 5} \right)}^2}} \;dt} = \hfill \\ \hfill \\ = - \frac{1}{2}\left\{ {\arcsin \left( {2x - 6} \right) + \left( {2x - 6} \right)\sqrt {1 - {{\left( {2x - 6} \right)}^2}} } \right\} \hfill \\ \end{gathered} \]

  49. Michele_Laino
    • one year ago
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    I'm pondering...

  50. Michele_Laino
    • one year ago
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    we have this result: the function: \[h'\left( x \right) = 4\sqrt {1 - {{\left( {\frac{{2x - 3}}{2}} \right)}^2}} \] is equal to zero at x=5/2 and x=1/2, furthermore it is always positive, so the function h(x) in that subintervals is always an increasing function

  51. Michele_Laino
    • one year ago
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    subinterval*

  52. Michele_Laino
    • one year ago
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    whereas, the function: \[h'\left( x \right) = - 2\sqrt {1 - {{\left( {2x - 6} \right)}^2}} \] is equal to zero at x=5/2 and at x=7/2. Furthermore it is always negative in that subinterval. So the function h(x) in that subinterval is a decreasing function

  53. Michele_Laino
    • one year ago
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    qualitatively, our function is represented by this drawing: |dw:1435090020580:dw|

  54. Michele_Laino
    • one year ago
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    namely x=5/2 is a local mximum for our function h(x)

  55. anonymous
    • one year ago
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    Thank you so, so much! This has helped me a ton!!

  56. Michele_Laino
    • one year ago
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    :)

  57. Michele_Laino
    • one year ago
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    more precisely, the second integral, is: \[h\left( x \right) = \int_4^{2x - 1} {f\left( t \right)\;dt} = - \int_4^{2x - 1} {\sqrt {1 - {{\left( {t - 5} \right)}^2}} \;dt} = \] and it was computed neglecting a constant term, which doesn't affect the value of the first derivative

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