First Derivative Test

a. Locate the critical ...

Question

First Derivative Test

a. Locate the critical points of f.

b. Use the First Derivative Test to locate the local maximum and minimum values.

c. Identify the absolute maximum and minimum values of the function on the given interval (when they exist).

f(x) = 2x⁵ - 5x⁴ - 10x³ + 4 on [-2,4]

Accepted Answer

Hello, in this video, we are going to be using the first derivative test and identifying the absolute maximum or minimum values. So, this problem tells us to consider the function F of X equal to 8/5 multiplied by X 5th, minus 2 X 4th, minus 32 X cubed plus 3 on the defined interval from -6 to 6. We want to find the critical points of F and use the first derivative test to classify these points. Then we want to determine the absolute maximum or minimum values of F's on the specified interval. Let's go ahead and start this problem by approaching the first part. The first part asks us to identify the critical points and to use the first derivative test to classify these points. No, so how do we start to find the critical points? Well, the critical points are located on the portions of F where the derivative or the slope is equal to 0. So in order to find the critical points, the first thing we need to do is take the derivative of the given function. So again, the function given to us is F of X equal to 8/5 multiplied by X 5th, minus 2X 4th, minus 32 X cubed plus 3. If we take the derivative with respect to x and use the power rule, the derivative F of X will equal to 8 x 4th power minus 8 x cubed minus 64 x 2. Now what we want to do is you want to set this derivative equal to 0 and solve for the values of X. If we solve for the values of X, these are going to give us the locations of the critical points. Now, before we factor or before we solve for X, notice that the terms in the derivative have at least one factor of 8 and one factor of X2. What we can go ahead and do is we can go ahead and factor out 8 X 2 from the derivative. That will allow us to rewrite the derivative as 8 X2, multiplied by X2 minus X minus 12, and we are setting this equal to 0. Notice that X2 minus X minus 12 is a factorable trinomial. We can further rewrite the derivative to be 8X2 multiplied by the quantity X minus 4, multiplied by the quantity X + 3. And we are setting this equal to 0. Now, by using the zero property, we can set each factor of X equal to 0 and solve for the independent values of X. That will give us 8 X2 equal to 0, X minus 4 equal to 0, and X + 3 equal to 0. If we divide by 8 in the first equation and take the square root of both sides, this will give us the value X is equal to 0. In the second equation, we just have to add 4 to both sides, and that will give us X is equal to 4. And in the final equation, subtracting 3 to both sides will give us X is equal to -3. What this means is that we have 3 critical points. For the values of X. We have three critical points located at X is equal to -3, 0, and 4, and this is going to be the first solution to this problem. So, we've identified the critical points of the graph of F. How can we use the first derivative test to classify these points? Well, the first derivative test will allow us to determine whether these points are absolute I mean local minimums or local maximums. Now, again, keep in mind that we are testing these points on the interval from -6 to 6. So the best way we can do this is we can go ahead and create a number line. Now, we are going to label our boundaries of the interval on the left and right hand side of the number line. We have -6 and 6. Now what we want to do is you want to plot our critical points on this number line. Again, our critical points are at -3. 0 and 4. Now, in order to determine the behaviors of the graph by the first derivative test, we are going to pick testing points in each of these intervals and determine whether we have a negative or positive value. If the value is positive, the behavior of the graph is increasing in that interval, and if the behavior is negative, then the behavior of the graph is decreasing on the interval. So let's go ahead and pick a testing point on the interval from -6 to -3. One testing point that we can pick is the value -4. Now what we want to do with this testing point is you want to go ahead and plug into the derivative. This will tell us the behaviors of the graph of either increase or decrease. So, if we take -4 and plug this into the derivative, F of -4 will give us the value of 1024. Since the value in this interval is positive, the region of the graph is increasing on the interval from -6 to -3. Now, let's go ahead and pick a testing point between -3 and 0. We can pick a testing 0.-1. F of -1 will give us the value of -80. Since the value is negative, that means the graph is decreasing in this interval. No Let's go ahead and pick a testing point on the interval from 0 to 4, and that testing point can be 1. F1 will give us the value of negative 96. Since this value is negative, the graph is decreasing from 0 to 4. Now, let's go ahead and pick a testing point between 4 and 6, and that testing point can be positive 5. F of 5 will give us the value of 1600, and since this is a positive value, the region of the graph is increasing from 4 to 6. Now, by the first derivative test, let's go ahead and take a look at the first critical point. Since the regions of the graph are increasing on the interval from -6 to -3, then decreasing from -3 to 0, by the first derivative test, this will be a local maximum. Now, since the intervals from -3 to 0 and from 0 to 4 are both decreasing, this does not classify as a local maximum or local minimum by the first interval test, or by the first derivative test. So this is neither. Now, what about 4? Well, since the graph is decreasing on the interval from 0 to 4, then increasing from 4 to 6, by the first derivative test, this classifies as a local minimum. So, what this means is that we have a local max. At the value, X is equal to -3. And we also have a local minimum at the value X is equal to 4. And this is going to be the 2nd solution to this problem. Now, let's go ahead and approach the second half of this problem. The second half asks us to determine the absolute maximums and minimum values of the given graph on the specified interval. How can we go ahead and do that? Well, in order to determine the absolute maximum or minimum values, what we want to do is we want to take our endpoints and all of the critical points, and plug them into our original function. Once we do that, we will go ahead and compare the values together, and we will use those values to determine the absolute maximum or minimum value. Now, again, the given function, the original function, is 8/5 multiplied by X 5th power, minus 2 X 4th, minus 32 X cubed plus 3. Let's go ahead and start by plugging in the first boundary of the defined interval. If we plug in -6 into the function, that is going to give us a value of -4000, 40,000, 593 divided by 5, which approximates to negative. 88,118.0. Next, if we plug in the critical 0.-3 into our function, this gives us the value of 1,588 divided by 5, which approximates to 316.2. If we plug in the critical 0.0 F of 0 just gives us the value of 3. Next, if we plug in the next critical point to the original function, F of 4 gives us the value of negative 4,593 divided by 5, which approximates to negative 918.6. And finally, if we plug in the value F of 6, which is the defined boundary. That is going to give us a value of 14,703 divided by 5, which approximates to 2,940.6. Now what we want to do is we want to go ahead and compare these values together to identify the absolute maximum or absolute minimum. Let's start with the absolute maximum. Now, amongst these 5 values, which one has the lowest output when you plug it into the original function? Well, that value is going to be -8,118, and this occurs at the value X is equal to -6. What this means is that we have an absolute maximum absolute minimum, I'm sorry. We have an absolute minimum. At X is equal to -6, with the respective value -6, 8,118.0. This is going to be the 2nd to final solution for the problem. And now for the absolute maximum, which one of these values is the highest possible value amongst the five we plugged in? Well, that value occurs at 2,940.6, with the respective value X is equal to 6. So we have an absolute maximum at the end point X is equal to 6 with the respective 0.6,2,940.6, and this is going to be the final solution to this problem. So, I hope this video helps you in understanding how to use the first derivative test and to find the absolute maximum minimum values, and I will go ahead and see you all in the next video.

First Derivative Test

a. Locate the critical points of f.
b. Use the First Derivative Test to locate the local maximum and minimum values.
c. Identify the absolute maximum and minimum values of the function on the given interval (when they exist).

f(x) = 2x⁵ - 5x⁴ - 10x³ + 4 on [-2,4]

Key Concepts

Critical Points