g is antideriv. of f = (x^2 + 4x)^(1/3); find g(1) / etc

[xc630]

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

 

[skeeter]

Re: some help please

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?

I do not believe this particular problem was meant to be done without using technology. Finding the antiderivative of f(x) is no simple task.

What needs to be done is setting up a definite integral and using the fundamental theorem of calculus ...

g(5) - g(1) = INT{1 to 5} f(x) dx

solving for g(1) ...

g(1) = g(5) - INT{1 to 5} f(x) dx

g(1) = 7 - INT{1 to 5} f(x) dx

... get out your calculator or use your integration program to evaluate the right side of the last equation.

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?

rate of production is A'(t) ... to find the maximum rate of production you have to analyze A'(t) by using A''(t).
A''(t) = -24(t-3) = 0 at t = 3.
Since A''(t) changes from (+) to (-) at t = 3, A'(t) is a maximum at t = 3. So, t = 3 is the time when the rate of production is increasing the fastest.


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 where did your negative sign in the exponent go?
0.159t= ln (3000/5500)
t= - 3.81
if thats right how is it negative