WEBVTT
00:00:00.640 --> 00:00:14.520
Find the local maxima and local minima of π of π₯ is equal to negative five π₯ squared over three plus two π₯ minus one-sixth times the natural logarithm of π₯, if any.
00:00:15.080 --> 00:00:18.040
Weβre given a function π of π₯.
00:00:18.840 --> 00:00:25.200
We need to determine all of the local maxima and local minima of this function if there are any.
00:00:26.040 --> 00:00:31.520
To find these points, we need to notice something about our function π of π₯.
00:00:32.440 --> 00:00:34.400
We know how to differentiate π of π₯.
00:00:35.080 --> 00:00:45.000
This is because π of π₯ is the sum of a polynomial and the natural logarithm function, and we know how to differentiate both of these.
00:00:45.680 --> 00:00:58.040
We also know that local extrema of a function will always occur when the derivative of our function is equal to zero or where the derivative does not exist.
00:00:58.800 --> 00:01:02.680
We also sometimes call these critical points.
00:01:03.040 --> 00:01:09.520
So we need to find an expression for π prime of π₯ to find all of our critical points.
00:01:10.160 --> 00:01:23.960
We know π prime of π₯ will be equal to the derivative of negative five π₯ squared over three plus two π₯ minus one-sixth times the natural logarithm of π₯ with respect to π₯.
00:01:24.880 --> 00:01:30.160
And we in fact know how to differentiate this term by term.
00:01:30.880 --> 00:01:37.040
The first two terms in this expression can be differentiated by using the power rule for differentiation.
00:01:37.720 --> 00:01:48.960
And to differentiate our third and final term, we need to remember the derivative of the natural logarithm with respect to π₯ is equal to the reciprocal function, one over π₯.
00:01:49.680 --> 00:01:53.800
So we can now evaluate this derivative.
00:01:54.440 --> 00:01:59.880
Again, we differentiate the first two terms by using the power rule for differentiation.
00:02:00.480 --> 00:02:06.000
We multiply by our exponent of π₯ and reduce this exponent by one.
00:02:06.520 --> 00:02:16.640
We get that π prime of π₯ is equal to negative 10π₯ divided by three plus two minus one divided by six π₯.
00:02:17.280 --> 00:02:21.320
We want to find all the critical points of our function π of π₯.
00:02:22.040 --> 00:02:31.440
Remember, thatβs where the derivative of π of π₯ is equal to zero or where the derivative does not exist.
00:02:31.680 --> 00:02:37.560
And we can see that π prime of π₯ is the sum of rational functions.
00:02:38.120 --> 00:02:45.040
And the only time a rational function will not be defined is when the denominator is equal to zero.
00:02:45.720 --> 00:02:50.200
And the only denominator here which can equal zero is six π₯.
00:02:51.120 --> 00:03:03.400
So π prime of π₯ will exist for all values of π₯ except when six π₯ is equal to zero, which is of course when π₯ is equal to zero.
00:03:03.400 --> 00:03:08.960
But before we carry on, thereβs one more thing we need to do.
00:03:09.720 --> 00:03:16.320
We need to check that π₯ is equal to zero is in the domain of our function π of π₯.
00:03:16.320 --> 00:03:26.840
This is because we canβt possibly check the derivative of our function at zero if our original function is not even defined when π₯ is equal to zero.
00:03:27.360 --> 00:03:33.840
And in fact we can see that π₯ is equal to zero is not in the domain of π of π₯.
00:03:34.240 --> 00:03:38.720
Otherwise, we would need to take the natural logarithm of zero.
00:03:39.200 --> 00:03:43.520
So zero is not a critical point of our function π of π₯.
00:03:44.120 --> 00:03:51.440
And in fact, this tells that π prime of π₯ is defined for all values of π₯ on the domain of π of π₯.
00:03:52.120 --> 00:03:59.880
So the only critical points of our function will be where the derivative of π of π₯ is equal to zero.
00:04:00.760 --> 00:04:05.840
One way of solving this equation is to multiply the entire equation through by π₯.
00:04:06.720 --> 00:04:11.520
Remember, weβve already explained that π₯ cannot be equal to zero.
00:04:12.320 --> 00:04:28.960
Multiplying through by π₯ and simplifying, we get that π prime of π₯ will be equal to zero when negative 10π₯ squared over three plus two π₯ minus one-sixth is equal to zero.
00:04:29.200 --> 00:04:42.920
And this is just a quadratic in π₯, so we could solve this by using a quadratic solver or the quadratic formula.
00:04:43.680 --> 00:04:49.680
Another way of solving this is to multiply our equation through by negative six.
00:04:50.120 --> 00:04:58.440
We get 20π₯ squared minus 12π₯ plus one is equal to zero.
00:04:59.400 --> 00:05:15.520
And then, either by inspection or the factor theorem or one of the methods we mentioned previously, we can factor this to give us 10π₯ minus one times two π₯ minus one is equal to zero.
00:05:15.760 --> 00:05:24.520
And we know if the product of two factors is equal to zero, then one of these two factors must be equal to zero.
00:05:24.960 --> 00:05:33.160
In other words, either 10π₯ minus one is equal to zero or two π₯ minus one is equal to zero.
00:05:33.600 --> 00:05:39.400
And then we can just solve both of these linear equations for π₯.
00:05:40.040 --> 00:05:44.960
We get either π₯ is equal to one over 10 or π₯ is equal to one-half.
00:05:46.400 --> 00:05:57.800
And just as we did before, we should also check that our function π of π₯ is defined when π₯ is equal to one over 10 and when π₯ is equal to one-half.
00:05:59.040 --> 00:06:10.840
And if we did this, we can see, of course, we can substitute π₯ is equal to one over 10 or π₯ is equal to one-half into the polynomial part of π of π₯.
00:06:11.600 --> 00:06:18.680
And both of these two values are positive, so we can also substitute them into the natural logarithm of π₯.
00:06:19.560 --> 00:06:24.880
So both of these values are in the domain of π of π₯.
00:06:25.520 --> 00:06:31.600
Therefore, both of these points are critical points of our function π of π₯.
00:06:32.680 --> 00:06:43.000
The derivative of π of π₯ at these points is equal to zero, which of course tells us that these will be local extrema of our function.
00:06:43.480 --> 00:06:54.920
But we still need to determine whether these are local maxima or local minima and we need to find their values.
00:06:55.920 --> 00:06:58.960
And thereβs a few different ways we could do this.
00:06:59.240 --> 00:07:03.640
For example, we could use the first derivative test.
00:07:04.280 --> 00:07:16.320
However, if we look at our expression for the first derivative of π of π₯ with respect to π₯, we see we can differentiate this one more time.
00:07:16.320 --> 00:07:21.040
This means we could also use the second derivative test.
00:07:22.120 --> 00:07:30.000
To use the second derivative test, weβre going to need to find an expression for the second derivative of π with respect to π₯.
00:07:30.800 --> 00:07:36.040
We can do this by differentiating π prime of π₯ with respect to π₯.
00:07:36.720 --> 00:07:45.400
Thatβs the derivative of negative 10π₯ over three plus two minus one over six π₯ with respect to π₯.
00:07:46.520 --> 00:07:57.120
And to make this easier to differentiate, weβll rewrite negative one over six π₯ as negative one-sixth multiplied by π₯ to the power of negative one.
00:07:57.120 --> 00:08:05.640
This means we can now evaluate this derivative term by term by using the power rule for differentiation.
00:08:06.200 --> 00:08:17.000
And then by applying the power rule of differentiation term by term, we get negative 10 over three plus one-sixth times π₯ to the power of negative two.
00:08:17.720 --> 00:08:26.680
And of course by using our laws of exponents, we can rewrite π₯ to the power of negative two in our denominator as π₯ squared.
00:08:27.360 --> 00:08:36.040
This gives us that π double prime of π₯ is equal to negative 10 over three plus one over six π₯ squared.
00:08:36.600 --> 00:08:41.720
We now need to recall the second derivative test for a function π of π₯.
00:08:42.200 --> 00:08:58.680
We recall this tells us if π₯ is a critical point of the function π and the second derivative of π with respect to π₯ at π₯ is positive, then it must be a local minima.
00:08:59.080 --> 00:09:10.920
However, if the second derivative of π with respect to π₯ at π₯ is negative, then it must be a local maxima.
00:09:10.920 --> 00:09:19.160
So we now need to evaluate the second derivative of π of π₯ with respect to π₯ at each of our critical points.
00:09:19.800 --> 00:09:22.400
Weβll start with π₯ is equal to one over 10.
00:09:22.960 --> 00:09:25.800
Weβll write this as 0.1.
00:09:26.360 --> 00:09:39.920
Substituting 0.1 into our expression for π double prime of π₯, we get negative 10 over three plus one divided by six times 0.1 squared.
00:09:40.400 --> 00:09:48.840
And if we evaluate this expression, we get 40 divided by three, and we need to notice this is positive.
00:09:49.280 --> 00:10:00.480
And because the second derivative of π of π₯ with respect to π₯ at 0.1 is positive, this means around this value of π₯, the slopes of our function is increasing.
00:10:01.240 --> 00:10:06.280
This means it must be a local minima.
00:10:07.280 --> 00:10:14.520
We can do exactly the same at our other critical point when π₯ is equal to one-half.
00:10:15.480 --> 00:10:18.560
Weβll write this as 0.5.
00:10:19.360 --> 00:10:24.640
We need to substitute 0.5 into our expression for π double prime of π₯.
00:10:25.480 --> 00:10:33.120
We get negative 10 over three plus one divided by six times 0.5 squared.
00:10:34.000 --> 00:10:43.640
This time, if we evaluate this expression, we get negative eight divided by three, which we need to notice is negative.
00:10:44.240 --> 00:10:51.480
So when π₯ is equal to 0.5, the second derivative of π of π₯ is negative.
00:10:52.280 --> 00:10:59.960
This means the slopes of our function are decreasing, which means we must be at a local maxima.
00:11:00.560 --> 00:11:04.080
But remember, thereβs one more thing we need to do.
00:11:04.560 --> 00:11:11.600
The question wants us to actually find the values of the local maxima and local minima of this function.
00:11:12.440 --> 00:11:19.920
To do this, all we need to do is substitute the values of our critical points into our function π of π₯.
00:11:20.920 --> 00:11:27.880
So letβs clear some space and find the values of our local maxima and local minima.
00:11:28.720 --> 00:11:32.040
Weβll start when π₯ is equal to 0.1.
00:11:32.800 --> 00:11:51.080
Substituting π₯ is equal to 0.1 into our function π of π₯, we get negative five times 0.1 squared divided by three plus two times 0.1 minus one-sixth times the natural logarithm of 0.1.
00:11:51.800 --> 00:12:04.440
And if we evaluate this expression, we can simplify it to get 11 divided by 60 minus one-sixth times the natural logarithm of one over 10.
00:12:05.240 --> 00:12:11.760
And we can do the same for our other critical point when π₯ is equal to one-half.
00:12:12.320 --> 00:12:20.560
Weβll write this as 0.5 and substitute π₯ is equal to 0.5 into our function π of π₯.
00:12:21.760 --> 00:12:35.080
We get negative five multiplied by 0.5 squared divided by three plus two times 0.5 minus one-sixth times the natural logarithm of 0.5.
00:12:35.880 --> 00:12:46.080
Then, evaluating and simplifying this expression, we get seven divided by 12 minus one-sixth multiplied by the natural logarithm of one-half.
00:12:46.800 --> 00:12:55.320
And remember, we already showed that this is a local maxima of our function, and this gives us our final answer.
00:12:56.200 --> 00:13:13.680
Therefore, we were able to show that the function π of π₯ is equal to negative five π₯ squared over three plus two π₯ minus one-sixth times the natural logarithm of π₯ only has two local extrema.
00:13:14.360 --> 00:13:25.200
It has a local minimum with a value of 11 over 60 minus one-sixth times the natural logarithm of one over 10 when π₯ is equal to one-tenth.
00:13:25.720 --> 00:13:36.080
And it has a local maximum value of seven over 12 minus one-sixth times the natural logarithm of one-half when π₯ is equal to one-half.