<p>For part d:
I found that the entries were being processed at t=12 most quickly(P(12)=8) not t=9.184(P(9.184)= 5.089?</p>
<p>Was I right?-t=12
(For max min...you have to check the end of the intervals)</p>
<p>FORM A BTW</p>
<p>My brain glitched and I wrote 10.837 something and thought that was the max, it was actually 9.184, I have no idea what i was doing lol</p>
<p>Wait…was t=12 the answer though…since the rate was a maximum at this time/end point …even though t=9.184 was a relative maximum too…</p>
<p>No, I think it was a quadratic or something, I forgot how it worked.</p>
<p>for that one you found that the critical point was a minimum using the first derivative test and a sign chart so according to the extreme value theorem, there must be a minimum and maximum on the closed interval [a,b[ so 12 would be ehwere it attained the max.</p>
<p>YAY…I was Right^^
Thanks for confirming my answer!</p>
<p>find when P’(t) = 0. This happens at x = 9.184 and 10.816. If you look at the graph, you see that the max is at 9.184 because it goes from positive to negative.</p>
<p>Alwy you forgot to check the ends of the interval…</p>
<p>t=12 is the answer</p>
<p>I hate my life</p>