00:01
We are going to use the first derivative test to find the location of all local strema in the interval 0 ,4 for the function f of x equals negative 3x to the fourth times natural logarithm of 3x.
00:16
If there is more than one local maximum, write each value of x separated by comma.
00:23
If a local maximum or local minimum does not occur in the interval, enter this symbol here, empty, in the appropriate box.
00:35
First, we know that on this interval 0 ,4 this function is continuous.
00:40
The key thing is that we are not including 0, so natural logarithm of 3x is well defined as 0.
00:50
The arguments 3x are positive for x on the interval 0 ,4, so natural logarithm of 3x is well defined and is continuous.
01:01
X to the fourth is a polynomial, it is continuous, negative 3 is a constant.
01:04
This function is continuous on 0 ,4 and is differentiable also.
01:09
We need to find the derivative to apply the first derivative test.
01:14
The first derivative of this function is equal to negative 3 is a constant times, we have a product x to the fourth times natural logarithm of 3x.
01:27
We apply the derivative of a product of two functions.
01:30
That is the derivative of the first factor, 4x cubed, that is the derivative of x to the fourth.
01:38
That times the second factor, natural logarithm of 3x, plus the first factor x to the fourth, times the derivative of the natural logarithm of 3x.
01:50
That is 1 over 3x times the derivative of 3x, which is 3.
01:56
Here we have applied the same rule.
01:58
This is negative 3 times 4x cubed, natural logarithm of 3x, plus, and then 3 here cancel out, and x to the fourth, one of the factors in x to the fourth, cancel with the x in the denominator, so we get x cubed.
02:24
So this is negative 3x cubed is a common factor, times 4 natural logarithm of 3x, plus 1.
02:41
So the derivative of this function is negative 3x cubed times 4 natural logarithm of 3x, plus 1.
03:02
Now we need to know which are the critical numbers of f, that is the values where the first derivative, values in the domain of f, where the derivative is 0 or does not exist.
03:18
In this case the derivative exists for all x on the interval 0, 4, so we look only for critical numbers where the derivative is 0.
03:31
Good, so the first derivative of f is 0, is equivalent to negative 3x cubed times 4 natural logarithm of 3x, plus 1 equals 0.
03:51
So this is the same as negative 3 is a constant, cannot be 0, x cubed is 0 when x equals 0, or we can have the other factor here, inside parenthesis, 4 natural logarithm of 3x, plus 1 equals 0.
04:09
But x equals 0 does not belong to the open interval 0, 4, so x equals 0 is not a critical number of this function on 0, 4.
04:35
So let's see what happens with the other equation, that is this one here.
04:41
Now, 4 natural logarithm of 3x, plus 1 equals 0, implies that 4, ok let's put it directly, natural logarithm of 3x, equals negative 1 fourth.
05:01
And so, taking exponential of both sides, we get 3x is going to be e to the negative 1 fourth, which is the same as 1 over e to the 1 fourth, which is the same as 1 over 4th root of e.
05:23
And so, solving for x, finally, x is equal to 1 over 3 times 4th root of e.
05:36
And this number is about, you have an idea, 0 .2596, so it's very close to 0 .26.
05:50
And this value is in the interval 0, 4, so it is a critical number of the function on that interval.
06:01
And indeed it's the only one, because the derivative exists for every point on 0, 4, and it is 0 only on this point inside the interval 0, 4...
