00:01
So for this question, we are asked to find a local max and the local min of the following function.
00:08
F of x equals 3 plus 6x squared minus 4x cubed.
00:14
And we want to use the first and second derivative tests in order to solve this problem.
00:21
So either way, we want to find the derivative.
00:26
So the derivative of 3 is 0, the derivative of 6x squared is 12x, and then the derivative of negative 4x cubed.
00:34
So the 3 comes down, which gives us negative 12, and then our new exponent's 3 minus 1, which is 2.
00:41
So we have 12x minus 12x squared.
00:43
Now, in order to find where our max or min might be, we have to find the critical points, right? so that's where the derivative either is equal to 0 or does not exist.
00:55
So our first derivative here is polynomial, so it exists everywhere.
00:59
So we need to figure out where this thing is, equals 0.
01:03
So let's set this equal to 0 and solve.
01:07
Notice that we can factor out a 12x, and we're left with 1 minus x and then the other factor.
01:17
And so basically from the first factor, 12x equals 0 gives us, right, so the only way that you know, two things multiplied by each other can be 0 is if 1 or both of them is 0, right? so the 12x piece could be 0, in which case, x is 0 or the 1 minus x factor could equal 0 in which case x equals 1.
01:43
So we have two critical points.
01:50
Now with these critical points we can do the first derivative test, which is a matter of determining if the derivative changes from positive to negative at the critical point or from negative to positive.
02:04
And the way that i like to do the first derivative test is graphically.
02:09
So the first derivative test, because you can really see what's going on with the first derivative test when you look at the graph.
02:15
So let's look at the graph of the derivative.
02:19
We just saw that the zeros were x equals 0 and x equals 1.
02:24
And this thing is a it's a quadratic, so it's going to be a parabola and the term in front of x squared is negative.
02:30
So it's a downward facing parabola.
02:34
So even just with this sketch, right, we have our critical point at x equals 0 and our point of x equals 1.
02:43
And if i focus in on x equals 0, notice that i go from negative values, right? if i'm like to the left of the critical point is a negative value, to the right of it's a positive value.
03:00
Right? so if i go from negative to positive, that means that the slopes on my original function, right? i don't even have to know what the graph looks like.
03:08
It just means that since the graph had a negative derivative, it means it had a negative slope on its tangent line.
03:16
So it was decreasing.
03:18
And then at this critical point, it switched to being a positive slope.
03:23
So if you're going from decreasing and you switch to being increasing, you are a minimum, a local minimum.
03:34
Now, on the other hand, if i focus on x equals 1, just to the left of that, i have a positive derivative, which means we had positive slopes and then we switch to negative derivative values.
03:47
So your slopes are negative, so you're decreasing...