xc630

02-10-2007, 10:39 PM

Hello I would like help in solving or at least starting these 3 problems

1) If fis the function defined by f(x)= (x^2+4x)^1/3 and g is an antiderivative of f such that g (5) = 7, then g (1) = ?

I do not understand how you can find the antiderative of an original function. Can't you only tkae the antiderative of a derative or second derivative?

2) The amount A(t) of a certain item produced in a factory is given by

A(t)= 4000 + 48(t-3) - 4(t-3)^3

where t is the number of hours of production since the beginning of the workday at 8:00 am. At what time is the rate of production increasing most rapidly?

For this one if I find the derivative and graph it out and say at 1 hour the rate is 0 while at 2 the rate is 36 and this turns out to be ther greatest increase between hours does that mean the rate is increasing most rapidly at t= 2 or somehwere in between t= 1 and t= 2?

3) A population grows according to the equation P(t) = 6000-5500e^-0.159t for t is equal or greater than 0, t measured in years. This popualtion will approach a limiting value as time goes on. During twhich year will the population reach half of this limiting value?

I found this limiting value to be 6000. Therefore I did

3000= 6000-5500e^-0.159t

5500e^-0.159t = 3000

e^0.159t= 3000/5500

0.159t= ln (3000/5500)

t= - 3.81

if thats right how is it negative

1) If fis the function defined by f(x)= (x^2+4x)^1/3 and g is an antiderivative of f such that g (5) = 7, then g (1) = ?

I do not understand how you can find the antiderative of an original function. Can't you only tkae the antiderative of a derative or second derivative?

2) The amount A(t) of a certain item produced in a factory is given by

A(t)= 4000 + 48(t-3) - 4(t-3)^3

where t is the number of hours of production since the beginning of the workday at 8:00 am. At what time is the rate of production increasing most rapidly?

For this one if I find the derivative and graph it out and say at 1 hour the rate is 0 while at 2 the rate is 36 and this turns out to be ther greatest increase between hours does that mean the rate is increasing most rapidly at t= 2 or somehwere in between t= 1 and t= 2?

3) A population grows according to the equation P(t) = 6000-5500e^-0.159t for t is equal or greater than 0, t measured in years. This popualtion will approach a limiting value as time goes on. During twhich year will the population reach half of this limiting value?

I found this limiting value to be 6000. Therefore I did

3000= 6000-5500e^-0.159t

5500e^-0.159t = 3000

e^0.159t= 3000/5500

0.159t= ln (3000/5500)

t= - 3.81

if thats right how is it negative