Just need someone to double check this problem. I just don't feel comfortable with my answer.

Question

Just need someone to double check this problem. I just don't feel comfortable with my answer.

An enviromental study of a certain communtiy suggests that t years from now, the average level of carbon monoxide in the air will be Q(t)=0.05t^2+0.01t+3.4 parts per million. (Read as .05t raised to the second power). By approximately how much will the carbon momoxide level change during the next 6 months?

Q'(t) = 0.1t +0.1
change in t=.5
Q(.5)=.3+0.005+3.4 =3.705
Q'(.5) =0.05+0.1=.15

Q(.5)=.5(.15)+3.705=3.78

Am I right? If not, where did I go wrong?

I would not do the deraviative.

change in six months = Q(.5) - Q(0)
= 3.705-3.4
=.305

Answer 1

Your calculation is correct. The change in the carbon monoxide level during the next 6 months is approximately 0.305 parts per million. You correctly used the given function Q(t) to calculate the carbon monoxide level at t = 0.5 (6 months) and at t = 0 (current level). Then, subtracting the current level from the level after 6 months gives you the change in the carbon monoxide level. Well done!