##### 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.

May 6th, 2015

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^2-15.9t+constant. at t=2

9.9=-18.2+constant

constant=28.1

f(x)= 3.4t^2-15.9t+28.1

May 7th, 2015

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.

May 7th, 2015

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^2-15.9t+28.1

f(x)=3.4(9)^2-15.9(9)+28.1

f(x)=106.4 billion for 2009

May 7th, 2015

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)

May 7th, 2015

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.

May 7th, 2015

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)

May 7th, 2015

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) = ?

May 7th, 2015

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).

May 7th, 2015

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May 6th, 2015
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May 6th, 2015
Dec 8th, 2016
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