WEBVTT
00:00:02.400 --> 00:00:08.690
Find the local maxima or minima of the function π of π₯ equals three π₯ to the fourth power minus two π₯ cubed.
00:00:10.250 --> 00:00:13.830
Local maxima and minima are examples of critical points for function.
00:00:14.020 --> 00:00:21.210
And we know that the critical points for function occur when its first derivative, π prime of π₯, is equal to zero or is undefined.
00:00:21.810 --> 00:00:25.190
Looking at our function π of π₯, we see that it is a polynomial.
00:00:25.330 --> 00:00:30.140
And we know that the derivative of a polynomial is defined for all values in its domain.
00:00:30.370 --> 00:00:34.460
So we donβt need to be concerned about π prime of π₯ being undefined.
00:00:34.590 --> 00:00:39.580
We only need to look for π prime of π₯ equal to zero to determine critical points.
00:00:40.890 --> 00:00:43.870
Letβs first find an expression for π prime of π₯ then.
00:00:43.870 --> 00:00:46.890
And we can do this using the power rule of differentiation.
00:00:47.430 --> 00:00:51.150
π prime of π₯ is equal to three multiplied by four π₯ cubed.
00:00:51.370 --> 00:01:03.880
Remember, we bring the exponent down and then reduce the exponent by one minus two multiplied by three π₯ squared, which simplifies to 12π₯ cubed minus six π₯ squared.
00:01:05.480 --> 00:01:12.350
To find the π₯-coordinates of critical points, we then set our first derivative equal to zero and solve the resulting equation for π₯.
00:01:12.750 --> 00:01:21.510
Now, we can take a common factor of six π₯ squared from each of these terms, giving six π₯ squared multiplied by two π₯ minus one is equal to zero.
00:01:22.840 --> 00:01:24.360
Six is not equal to zero.
00:01:24.390 --> 00:01:34.520
So setting the remaining two factors in turn equal to zero, we have π₯ squared equals zero, leading to π₯ equals zero and two π₯ minus one equals zero, leading to π₯ equals one-half.
00:01:34.970 --> 00:01:41.210
Weβve therefore found that this function π of π₯ has critical points at π₯ equals zero and π₯ equals one-half.
00:01:42.680 --> 00:01:46.670
We then need to evaluate the function itself at each of the critical points.
00:01:46.880 --> 00:01:56.030
Firstly, substituting π₯ equals zero gives π of zero equals three multiplied by zero to the fourth power minus two multiplied by zero cubed, which is equal to zero.
00:01:57.350 --> 00:02:04.830
Substituting π₯ equals one-half gives three multiplied by one-half to the fourth power minus two multiplied by one-half cubed.
00:02:06.450 --> 00:02:16.900
Thatβs three over 16 minus one over four or one-quarter, which we can think of as three over 16 minus four over 16, which is negative one over 16.
00:02:18.370 --> 00:02:24.680
The critical points of this function then occur at zero, zero and one-half, negative one over 16.
00:02:24.850 --> 00:02:27.960
But we donβt yet know what type of critical points there are.
00:02:28.220 --> 00:02:29.830
Thereβre three possibilities.
00:02:30.960 --> 00:02:34.600
They could be local maxima, local minima, or points of inflection.
00:02:35.060 --> 00:02:40.250
In order to classify each of our critical points, we need to use the second derivative test.
00:02:40.590 --> 00:02:45.950
The sign of the second derivative of our function tells us which type of critical point we have.
00:02:47.420 --> 00:02:54.650
If the second derivative of our function is negative, then this means that the first derivative or slope of the function is decreasing.
00:02:54.910 --> 00:02:56.770
And so we have a local maxima.
00:02:58.190 --> 00:03:04.450
If the second derivative is positive, then the slope or first derivative of the function π prime of π₯ is increasing.
00:03:04.630 --> 00:03:06.480
And so we have a local minimum.
00:03:07.860 --> 00:03:13.280
If the second derivative is equal to zero, then the critical point could be a point of inflection.
00:03:13.470 --> 00:03:14.910
But this isnβt sufficient.
00:03:15.000 --> 00:03:20.760
It is possible to have a local minimum or a local maximum when the second derivative is equal to zero.
00:03:21.040 --> 00:03:27.510
So if we find the second derivative is equal to zero, we need to perform other checks in order to classify the critical point.
00:03:29.220 --> 00:03:33.310
Letβs begin then by finding an expression for the second derivative of our function.
00:03:33.470 --> 00:03:40.760
And to do this, we need to differentiate the expression we found for the first derivative, which remember was 12π₯ cubed minus six π₯ squared.
00:03:41.230 --> 00:03:47.430
Differentiating gives 12 multiplied by three π₯ squared minus six multiplied by two π₯.
00:03:49.460 --> 00:03:53.190
That simplifies to 36π₯ squared minus 12π₯.
00:03:54.630 --> 00:03:57.780
Letβs check the critical point when π₯ is a half first of all.
00:03:58.020 --> 00:04:06.360
Substituting π₯ equals one-half into our second derivative gives 36 multiplied by one-half squared minus 12 multiplied by one-half.
00:04:07.820 --> 00:04:14.080
Thatβs 36 over four minus 12 over two or nine minus six, which is equal to three.
00:04:15.470 --> 00:04:21.400
Weβre not particularly interested in the value of the second derivative other than to notice that it is greater than zero.
00:04:21.560 --> 00:04:26.510
And so, by the second derivative test, we find that this critical point is a local minimum.
00:04:28.200 --> 00:04:35.120
When we substitute π₯ equals zero for our other critical point, we find that the second derivative is also equal to zero.
00:04:35.370 --> 00:04:40.490
And so here, the second derivative test hasnβt helped us with classifying this critical point.
00:04:40.740 --> 00:04:42.680
We need to perform a different check.
00:04:42.910 --> 00:04:45.000
And weβll use the first derivative test.
00:04:46.200 --> 00:04:50.580
What weβll do is check the sign of the first derivative either side of our critical point.
00:04:50.760 --> 00:04:54.070
So we can consider the shape of the curve around the critical point.
00:04:54.420 --> 00:05:00.630
Now, as our next critical point occurs when π₯ equals one-half, we need to choose π₯-values quite close to zero.
00:05:00.830 --> 00:05:04.220
So weβll choose π₯-values of negative one-quarter and one-quarter.
00:05:04.680 --> 00:05:11.950
We already know that π prime of π₯, the first derivative, is equal to zero when π₯ is equal to zero, at the critical point itself.
00:05:13.410 --> 00:05:25.910
Substituting π₯ equals negative one-quarter into our first derivative, which remember was 12π₯ cubed minus six π₯ squared, we find that the first derivative is equal to negative nine over 16 at this point.
00:05:26.380 --> 00:05:28.460
Weβre not particularly interested in the actual value.
00:05:28.650 --> 00:05:31.730
But we are interested in the fact that it is a negative value.
00:05:33.240 --> 00:05:43.780
Substituting π₯ equals positive a quarter into our first derivative gives 12 multiplied by one-quarter cubed minus six multiplied by one-quarter squared, which is negative three over 16.
00:05:44.040 --> 00:05:48.280
And again, what weβre interested in is the fact that the first derivative is negative here.
00:05:49.640 --> 00:05:53.010
This helps us to picture the shape of the curve at this critical point.
00:05:53.170 --> 00:05:58.220
If the slope is negative, then zero, then negative again, we can sketch the shape.
00:05:58.250 --> 00:06:01.880
And we see that this critical point is in fact a point of inflection.
00:06:03.180 --> 00:06:11.870
We only need to perform this first derivative test when the second derivative test fails to distinguish between a local minimum, a local maximum, or a point of inflection.
00:06:12.190 --> 00:06:16.430
But actually, the first derivative test is an acceptable method in its own right.
00:06:17.940 --> 00:06:21.230
The question only asked us about local maxima or minima.
00:06:21.390 --> 00:06:27.460
So we donβt need to mention the point of inflection in our final answer although we did need to check it in our working.
00:06:27.950 --> 00:06:34.510
We can conclude then that the function π of π₯ has a local minimum point at one-half, negative one sixteenth.