## anonymous 4 years ago 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.

1. anonymous

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

2. JamesJ

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.

3. JamesJ

talk to me

4. anonymous

hold on. im processing this information. lol

5. anonymous

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

6. JamesJ

5d + 5 I agree with. Calculate F'(1) again.

7. anonymous

idk what im doing wrong

8. anonymous

10s+5=1? no?

9. JamesJ

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

10. JamesJ

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

11. anonymous

15

12. JamesJ

yes

13. JamesJ

Thus 5d + 5 = 15

14. anonymous

i wasnt plugging in i was setting equal to 1

15. anonymous

16. JamesJ

Would seem so!

17. anonymous

yeah, it is