Homework help needed.
Calculus

Tutor: None Selected  Time limit: 1 Day 
1) Find the cost function if the marginal cost function is given by C'(x)= x^(3/4) + 5 and 16 units cost $212.
2) The approximate rate of change in the number (in billions) of monthly text messages is given by the equation f'(t) = 6.8t 15.9 where t represents the number of years since 2000. In 2002 (t=2) there were approximately 9.9 billion monthly text messages. Find the function that gives the total number (in billions) of monthly text messages in year t.
1)integreting C'(x)=x^(3/4)+5
C(x)=(7/4)*(x)^(7/4)+5(x)+constant 16 units, then $212
212=304+constant
constant=92
cost function=(7/4)*(x)^(7/4)+5(x)92
2)integrate f'(x) to get 3.4t^215.9t+constant. at t=2
9.9=18.2+constant
constant=28.1
f(x)= 3.4t^215.9t+28.1
The first question was incorrect, but the second was correct.
For the first one, it gave me the answer as being (4x^(7/4))/7 +5x + 412/7.
So, then, a similar question I was given for the first is: Find the cost function if the marginal cost function is given by C'(x) = x^(1/2)+9
And for the second, there was a second part: According to this function, how many monthly text messages were there in 2009?
I will review after these are completed. :) Thank you in advance.
1)integreting C'(x)=x^(3/4)+5
C(x)=(4/7)*(x)^(7/4)+5(x)+constant 16 units, then $212
212=(1072/7)+constant
constant=212(1072/7)
constant= 412/7
cost function=(4/7)*[(x)^(7/4)]+5(x)(412/7) is the correction of number one.
for the extras c'(x)=x^(1/2)+9
c(x)=(2/3)*[(x^(3/2)]+9x+constant.
2) the time t=9 upto 2009
f(x)= 3.4t^215.9t+28.1
f(x)=3.4(9)^215.9(9)+28.1
f(x)=106.4 billion for 2009
for the extras c'(x)=x^(1/2)+9
c(x)=(2/3)*[(x^(3/2)]+9x+constant.
212=(2/3)*[(16)^(3/2)]+9(16)+constant
constant= 76/3
c(x)=(2/3)*[(x^(3/2)]+9x+(76/3)
f(x)=106.4 billion for 2009  That was incorrect. Perhaps you mistyped, but it was 160.4, which I solved.
c(x)=(2/3)*[(x^(3/2)]+9x+(76/3) is incorrect, also.
F(X)=160.4 it was a mistype.
 for c'(x)=x^(1/2)+9
c(x)=[2x^(3/2)]/3 + 9x+c
212=[2(16)^(3/2)]/3 + 9(16) + c
212=128/3 + 144 + c
212(560/3)=c
c=76/3
c(x)=[2x^(3/2)]/3 + 9x+(76/3) which is a similar equation to (2/3)*[(x^(3/2)]+9x+(76/3)
Again, no. The correct answer that they gave was: [2x^(3/2)]/3 + 9x + (2/3)
A similar exercise to that one again: C'(x) = x^(2/5) + 9 and 32 units cost $410.
C(x) = ?
for every C'(x) integrate to find C(x). then use the condition 32units=$410 to find the constant of integration,c.
replace c to the equation of c(x).
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