<p>Mean Value Theorom does work ,it is continuous, just not differnetiable</p>
<p>Mean value theorem does not apply because the function was NOT differentiable at 8 and 20, but it WAS continuous.</p>
<p>"Mean Value Theorem: Let f(x) be differentiable on the open interval (a, b) and continuous on the closed interval [a, b]"</p>
<p>porsche: i think it was the points of 4 and 16 that weren't differentiable.</p>
<p>at 16, v(t) was 20.</p>
<p>that's different from v(20).</p>
<p>i made that mistake first, and went back and fixed it. but hey i don't know i could be wrong.</p>
<p>IT was continuos at 20.</p>
<p>not continous at (16, 20) v(16) was not continuous.</p>
<p>V(20) was continous though.</p>
<p>But isn't there aa point c that has the same slope as between the points a and b?</p>
<p>dru: oh, well whatever points they gave u to consider is what i used i think.
Sx91: you tell em lol</p>
<p>no, it wasnt differentiable. mean value theorem defuinitely doesnt apply</p>
<p>Yeah I'm almost positive that V(20) was continuous and you could just do the slope there.</p>
<p>has to be differntialbe and continuous..</p>
<p>first derivative of V(20) was definately continous..it was the middle point on the negative slope (a straight line)</p>
<p>it was V(16) = 20 that was the little CUSP, differnt side limit, w/e.</p>
<p>you can even look at the graph itself, the only slopes were like 5, 0, and -5... so obviously a slope btwn those values wont exist if its 3 linear functions for the graph...</p>
<p>it definitely doesnt work..... the average value was -5/6 but there was no value corresponding to that on the graph... it was either 0 or -5/2</p>
<p>sure it was differentialble at v'(20)... but its talking about the WHOLE interval, and it wasn't at v'(16)</p>
<p>About polar, for 3rd part :what happens to the graph between pi/something and pi/something, what did u put</p>
<p>It was -5/6
Does someone have the damn book to check
We were supposed to get them today</p>
<p>yeh i think i got wat BCgoUSC got...</p>
<p>Howd you find T(7) and what was the meaning the integral of the derivative of the T. And what number did you get for that..</p>
<p>Does anyone remember the second to last step for the last question of taylor? Was it something like |[(x-2)^2]/3]| < 1 ? I know i got that wrong but i just wanted to know how people got the interval to be [-1,5).</p>
<p>Come on people, answers about polar</p>
<p>I got the interval bracket -1, 5) anyone else, please confirm!</p>
<p>For polars, I think that the max was at pi/3.</p>
<p>Also, on the interval pi/3 to 2pi/3, it was decreasing, and I put it meant that the curve was going toward the pole.</p>
<p>was the area of polar = int 1/2 [ r(theta) ^2 dtheta] from 0 to pi ?</p>
<p>that's what I got.</p>
<p>Here's a clearer explanation for those of you who can't think in images:</p>
<p><a href="http://img31.echo.cx/img31/2971/bc1ox.jpg%5B/url%5D">http://img31.echo.cx/img31/2971/bc1ox.jpg</a></p>
<p>Now tell me I'm wrong.</p>
<p>Metra and yagun, I got your answers for the taylor question.</p>