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Math Help - Students in calculus course..

  1. #1
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    Question Students in calculus course..

    Students in calculus course were given a exam and each month thereafter they took an equivalent exam.The class average on the exam taken after t months is F(t)=86-6 ln(t+1).
    How many times has an equivalent exam been given if the average score is decreasing at a rate of one point per month.
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  2. #2
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    Quote Originally Posted by mathcalculushelp View Post
    Students in calculus course were given a exam and each month thereafter they took an equivalent exam.The class average on the exam taken after t months is F(t)=86-6 ln(t+1).
    How many times has an equivalent exam been given if the average score is decreasing at a rate of one point per month.
    You are being asked to find the solution of:

    F'(t)=-1

    CB
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  3. #3
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    Question

    F(t)=86-6ln(t+1)
    = 86- 6 ln 2
    = 86-4.15
    = 81.8 ...??

    is this what we are supposed to do?

    or like shall i be plugging in f(-1) ?
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  4. #4
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    You solve for -1, because the function gives the rate of change for the grades, in this case -1, when you solve for  t in this instance it will tell you how many months has passed and according to the problem one test a month
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    Quote Originally Posted by mathcalculushelp View Post
    F(t)=86-6ln(t+1)
    = 86- 6 ln 2
    = 86-4.15
    = 81.8 ...??

    is this what we are supposed to do?

    or like shall i be plugging in f(-1) ?
    Differentiate F(t) with respect to t (which gives the rate of decrease in scores per month), then set that equal to -1 and solve for t (because that is what is asked for).

    CB
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  6. #6
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    Question is this correct?

    Alright ..

    So here is what I have done
    f(t)=86-6ln(t+1)
    f'(t)= 6 (1/t+1)
    now f'(t)=-1

    6/(t+1)=-1
    6=-t-1
    6+1=-t
    -7=t

    Is this correct?
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  7. #7
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    Quote Originally Posted by mathcalculushelp View Post
    Alright ..

    So here is what I have done
    f(t)=86-6ln(t+1)
    f'(t)= 6 (1/t+1)
    now f'(t)=-1

    6/(t+1)=-1
    6=-t-1
    6+1=-t
    -7=t

    Is this correct?
    Does a negative value of t seem reasonable to you? If not, then perhaps you should check your work to see if you've made a careless mistake somewhere.
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