00:01
In this problem, we are beginning to understand the behavior and applications of logarithmic and exponential functions.
00:09
So let's first review the function that we're given.
00:11
We have q of t equals 135 times 1 minus e raised to negative 0 .05t plus 60.
00:22
So the first that we need to do is find the words per minute prior to taking this course.
00:28
So let's first take the derivative, q prime of t.
00:33
Our q prime of t is 135 times 0 .05e raised to negative 0 .05t.
00:41
Well, what does that mean? that means our first derivative is positive for the interval that we're given.
00:48
0 is less than or equal to t, which is less than or equal to 20.
00:53
So q prime of t is an increasing function.
00:57
So now let's take our second derivative.
01:01
Double prime, q double prime of t equals negative 135 times 0 .05 squared times e raised to negative 0 .05 t...