Question

C(0) = 1000 and C'(x) = 400e^(-0.1q) + 8. Then C(x) is A. -4000e^(-0.1q) + 8q...

C(0) = 1000 and C'(x) = 400e^(-0.1q) + 8. Then C(x) is

A. -4000e^(-0.1q) + 8q + 1000

B. -4000e^(-0.1q) + 8q + 5000

Notice that both A and B satisfy the 2nd condition: derivative of A or B = -4000e^(-0.1q)(-0.1) + 8 + 0 = 400e^(-0.1q) + 8.

But which one satisfies the 1st condition also?
Math   Calculus   
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Homework Answers

Answer #1

Given that C(0) = 1000

and C'(x) = 400 + 8

let say,

A(q) = -4000 + 8q + 1000

B(q) =  -4000 + 8q + 5000

Now,

A'(q) = -4000 (-0.1) + 8 + 0 = 400 + 8

B'(q) = -4000 (-0.1) + 8 + 0 = 400 + 8

A(0) = -4000 + 8 0 +1000 = -4000+1000 = -3000

B(0) = -4000 + 8 0 +5000 = -4000 + 5000 = 1000

so, we have B(0) = 1000

Thus, only B satisfies the 1st condition

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