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Homework Help: 3 free response practice AP problems

  1. Mar 30, 2005 #1
    Hi everyone. I need some help with these 3 problems. Any help would be appreciated.

    4. Let f be the function defined by f(x)= ln(2+ sin x) for pi <= x <= 2pi

    (a) Find the absolute maximum value and the absolute minimum value of f. Show your analysis that leads to your conclusion.

    (b) Find the x-coordinate of each point on the graph of f. Justify your answer.

    My work:

    f'(x) = (cos x) / (sin x +2)
    0= (cos x) / (sin x +2)
    0= cos x
    x= pi/2

    f''(x) = [ (sin x +2)(-sin x ) - (cos x)(cos x) ] / [ (sin x +2)^2 ]
    0= (sin x +2)(-sinx) - (cos x)(cos x)

    Now here is my problem. How do I solve that??

    5. The figure above shows the graph of f', the derivative of a function f. The domain of f is the set of all such that 0<x<2

    (a) Write an expression for f'(x) in terms of x
    (b) Given that f(1)=, write an expression for f(x) in terms of x
    (c) In the xy-plane provided below, sketch the graph of y=f(x)

    My Work:

    a. for (1,1) and (2,0)


    f'(x) = {x 0<= x <= 1
    {-x+2 1<= x <=2

    b. 0= .5(1^2) +c
    -.5 = c

    0= 2(1) -.5(1^2) +c
    0=2-.5 +c

    f(x)= {.5x^2 + -.5 0<= x <=1
    .........{2x - .5x^2 -1.5 1<= x <2

    6. Let P(t) represent the number of wolves in a population at time t years, when t>= 0. The population P(t) is increasing as a rate directly proportional to 800-P(y), where the constant of proportionality is k.

    (a) If P(0)= 500, find P(t) in terms of t and k
    (b) If P(2)=700, find k
    (c) Find lim P(t)

    i have no clue what to do for this one. since it says "rate" it must use related rates but how would you do it??

  2. jcsd
  3. Mar 30, 2005 #2
    #4, to find the absolute maximum and minimum of a graph, only the first derivative is required. The second derivative is used to find curvature.

    Solving the first derivative will give you the points where a critical point occurs. From there, you can look at the graph to see if it is a maximum or minimum. Don't forget to evaluate the endpoints of the interval.

    #5. I dont see a question

    #6. Thats just straight up confusing.
  4. Mar 30, 2005 #3
  5. Mar 30, 2005 #4
    On #5. Theres still alot of blanks in the question, you might want to re write it but
    a) the second part of the composite function is 1-x, not 2-x. You are starting from y = 1, not 2.

    b) i dontk now what f(1) is, but it looks like you have the right idea.

    c) Can't see

    6. That question doesnt make sense, did you write it correctly?
  6. Mar 30, 2005 #5
    For number 5 part b they did not give me a picture of the "xy-plane provided below"

    For number 4 would I have to solve the equation 0= (sin x +2)(-sinx) - (cos x)(cos x) to get the points of inflection?

    Yeah I wrote number 6 correctly
  7. Mar 30, 2005 #6
    Solving f''(x) will give you the concavic tendencies of the curve.

    #6 doesnt make sense to me.
  8. Mar 30, 2005 #7
    Yeah I don't understand number 6 either.

    How would I find the points of inflection for number 4?
  9. Mar 30, 2005 #8
    4. Let f be the function defined by f(x)= ln(2+ sin x) for pi <= x <= 2pi

    (a) Find the absolute maximum value and the absolute minimum value of f. Show your analysis that leads to your conclusion.

    [tex] f(x)= ln(2+ sin x) [/tex] and condition [tex] \pi <= x <= 2\pi [/tex]

    [tex]f'(x) = \frac{cosx}{sinx+2}[/tex]

    [tex]\frac{cosx}{sinx+2} = 0[/tex]

    [tex] cosx = 0, x = \frac{3\pi}{2}[/tex]

    There is a critical point at x = [tex] \frac{3\pi}{2} [/tex]

    You could look at a graph of the function to see if it is a minimum or maximum, but if you can't do that, then take a point to the left and right of it and evaluate the function there.
    [tex]f'(x) = \frac{cosx}{sinx+2}[/tex]

    Notice f'(x) for [tex] \pi < x < 3\pi/2 [/tex] is always negative.
    Notice f'(x) for [tex] 3\pi/2 < x < 2pi [/tex] is always positive.

    The curve is decreasing up to x = 3pi/2, and increasing from 3pi/2 to 2pi. You can conclude that x = 3pi/2 is the absolute minimum of the graph.

    Evaluating critical points & endpoints:
    [tex] f(3\pi/2) = ln(2+sin(3\pi/2)) = ln(2-1) = ln(1) = 0 [/tex]

    [tex] f(\pi) = ln(2+sin(pi)) = ln(2) [/tex]

    [tex] f(2\pi) = ln(2+sin(2\pi)) = ln(2) [/tex]

    The local minimum is at x = 3pi/2, and the value is 0. The local maximums are at x = pi, 2pi and have value ln(2).
  10. Mar 30, 2005 #9
    But would those points be the same thing as the points of inflection?
  11. Mar 30, 2005 #10
    It's not asking you for points of inflection. For those you would need to solve f''(x) = 0.
  12. Mar 30, 2005 #11
    How would you find part b of number 4?
  13. Mar 30, 2005 #12
    Thats a starngely asked question but:

    Part a) min f = 0, max f = ln2
    Part b) min x = 3pi/2, max x = pi,2pi
  14. Mar 30, 2005 #13
    would the answers to part a be : min x = 3pi/2, max x = pi,2pi because part a is asking you for the min and max?
  15. Mar 30, 2005 #14
    Oh my gosh I'm so sorry. For number 4 part b is suppose to read :
    (b) Find the x-coordinate of each inflection point on the graph of f. Justify your answer.
  16. Mar 30, 2005 #15
    For part b solve f''(x) = 0. The solutions to that equation are the points of inflection.
  17. Mar 30, 2005 #16
    f''(x) = [ (sin x +2)(-sin x ) - (cos x)(cos x) ] / [ (sin x +2)^2 ]
    0= (sin x +2)(-sinx) - (cos x)(cos x)

    but how would i solve that?
  18. Mar 31, 2005 #17
    [tex] \frac{-(sinx+2)sinx-cos^2x}{(sinx+2)^2} = 0[/tex]

    [tex] \frac{-sin^2x-2sinx-cos^2x}{(sinx+2)^2} = 0 [/tex]

    [tex] \frac{-(sin^2x+cos^2x+2sinx)}{(sinx+2)^2} = 0 [/tex]

    [tex] sin^2x+cos^2x = 1 [/tex]

    [tex] \frac{-2sinx-1}{(sinx+2)^2} = 0 [/tex]

    You should be able to go from there.
  19. Mar 31, 2005 #18
    ok I got -pi/6 as my answer. what would the answer be in order to fit in the domain given?
  20. Mar 31, 2005 #19
    The sin function is periodical by 2pi

    [tex] sin(x) = sin(2n\pi+x) [/tex]

    There is an integer value n where the function will fall in the domain.
  21. Mar 31, 2005 #20
    ok I understand now. Thank you so much for helping me :)!

    I have 1 more question.

    I think I got number 6 now. I got p(t) = -300e^(-kt) + 800
    and part c says find:

    lim P(t)

    how would I do this?
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