# Students in calculus course..

• November 5th 2009, 02:02 PM
mathcalculushelp
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.
• November 6th 2009, 12:01 PM
CaptainBlack
Quote:

Originally Posted by mathcalculushelp
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
• November 18th 2009, 01:09 PM
mathcalculushelp
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) ?
• November 18th 2009, 01:49 PM
RockHard
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
• November 18th 2009, 03:02 PM
CaptainBlack
Quote:

Originally Posted by mathcalculushelp
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
• November 18th 2009, 03:49 PM
mathcalculushelp
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?
• November 18th 2009, 05:58 PM
mr fantastic
Quote:

Originally Posted by mathcalculushelp
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.