Question

Let F(s)=5s^2+5s+4. Find a value of d greater than 0 such that the average rate of change of F(s) from 0 to d equals instantaneous rate of change of F(s) at s=1.

Answer 1

this is a similar prob to what i posted earlier but im getting the wrong answer so im doing something wrong

Answer 2

The average rate of change of F(s) from 0 to d is the average of F'(s) from 0 to d, which is
\[ \frac{1}{d} \int_0^d dF/ds \ ds = \frac{F(d) - F(0)}{d} \]

Hence you want to find a d such that
\[ \frac{F(d) - F(0)}{d} = F'(1) \]

So now you just need to evaluate.

Answer 3

talk to me

Answer 4

hold on. im processing this information. lol

Answer 5

so 5d+5=-2/5 ???

Answer 6

5d + 5 I agree with.

Calculate F'(1) again.

Answer 7

idk what im doing wrong

Answer 8

10s+5=1? no?

Answer 9

No, you want the value of dF/ds when s = 1. That's what F'(1) means.

Answer 10

F'(s) = 10s + 5. Hence F'(1) = ...?

Answer 11

15

Answer 12

yes

Answer 13

Thus
5d + 5 = 15

Answer 14

i wasnt plugging in i was setting equal to 1

Answer 15

so the answer is 2

Answer 16

Would seem so!

Answer 17

yeah, it is