<p>I was doing the 1998 free response questions, and I was having some trouble with the second one, which asked find all the absolute min or max of f(x)=2x*e^(2x). I had no problems taking the derivative and establishing that the relative min was -1/2 however, my question is how would I then go onto prove that it was the absolute minimum. Thanks!</p>
<p>use the second derivative test!!! If the second derivative is positive then the graph is concave up in that interval, if the graph is concave up, then the graph goes from decreasing to 0 to positive; this means that its a minimum. If the 2nd derivative is negative, then the graph is concave down; so the value is a max...</p>
<p>hope this helps :)</p>
<p>I used the first derivative test and I got the value was a min. so now knowing that is a min, what step do I have to take further to establish it as the ABSOLUTE minimum.</p>
<p>You test the end values and the critical points (the relative max's and mins) by plugging in those points in the original function and then compare. </p>
<p>If there are no endpts, changes are there's a property of the function that allows you to determine the absolute extrema (i.e. f' is negative on the interval before the point, positive afterwards).</p>
<p>^oh, lol, I misunderstood what you were asking...yeah do what snipez said, he's right..you plug in the endpoints of the interval, and the critical points, the smallest value of f will be the absolute min, and largest the max</p>
<p>Sorry to keep asking questions, but in this question no intervals are provided, so I am not sure what I would do...</p>
<p>^well, if you have NO intervals, and only ONE critical point, and ask for the absolute maximum, then usually once you establish that it is in fact a min, you're done. I mean there is no other technique other than presenting a graph of the equation along side the value to further prove it. The proving is done--no interval, one critical point, proved that critical point is a min--DONE..there is enough proof to conclude that at x=-0.5, you have an absolute min</p>
<p>also -0.5 is NOT the minimum, -0.5 is the x-value of the min, you need to plug -0.5 in the main f function to get the absolute min.</p>
<p>^ok i got it thanks, and ya i was just writing down my x coordinate the whole time.</p>
<p>"also -0.5 is NOT the minimum, -0.5 is the x-value of the min, you need to plug -0.5 in the main f function to get the absolute min."</p>
<p>damn, thanks for reminding me... THAT would have been bad</p>
<p>^lol, I know, I've left the answer as x=whatever on soooo many practice tests that now I'm almost paranoid about it...another thing i do is always forget to put the dx's at the end of each integrand :)</p>