Graphs and Graphing

Graph th...

Question

Graphs and Graphing

Graph the curves in Exercises 33–42.

y = 𝓍³ (8―𝓍 )

Accepted Answer

Hello. In this video, we are going to be sketching the graph of Y is equal to 10 X cubed multiplied by 12 minus X, given that the first derivative is Y equal to 2/5X2 multiplied by 9 minus X, and the second derivative Y is equal to 65 X multiplied by 6 minus X. Now, we are going to start this problem by working with the base function Y. Using the base function, the first thing that we can determine is the domain of the function. Now, if we look at the given function, the function is in the form of a polynomial. Since the function is in the form of a polynomial, then the domain is going to be all real numbers from negative infinity to positive infinity. The next thing we are going to do is we are going to analyze the graph's symmetry. Now, in order to determine the symmetry of the graph, we are going to plug in the value of negative X into the function. There are two outputs that we want to keep an eye out for. First, if we plug in negative X and we get out the original function F of X, then there is going to be symmetry about the about the Y axis, therefore, the function is even. And if we plug in negative X and we get negative F of X, then there is going to be symmetry about the X axis, therefore, the function is odd. So, if we plug in negative X into the original function, we get Y is equal to 1/10 multiplied by negative X cubed, multiplied by 12 minus negative X. X cubed will simplified to be negative X, and multiplying that to 1/10 will give us negative 1/10 X cubed. On the second portion of the function, negative negative X will be positive, so we are left with 12 + X. Now, right away we can see that when we plug in negative X into the function, we do not get the original function. So the function is not going to be even. Furthermore, if we were to factor out a -1 from 12 + X, we will not get a negative version of the original function. So that rules out the possibility of the function being odd. Therefore, there is no symmetry for the function. Now that we've discussed symmetry, we can use the original function to determine the asymptotes, or uh the X and Y intercept. So, let's go ahead and determine the X intercepts. In order to determine the X intercepts, we are going to solve for the values of X that cause the function to be zero. Or in other words, we have 1/10 X cubed, multiplied by 12 minus X, and this should equal to 0. By using the zero property, we can set each factor of X equal to 0. 10 X cubed is going to be equal to 0. And simplifying this will leave us with X's equal to 0, which is going to be the X intercept we're looking for. Furthermore, 12 minus X is going to equal to 0, and solving for X will give us X's equal to 12. So we have two X intercepts, one at the origin and one at 120. Furthermore, because we have an X intercept at the origin, that also tells us that the Y intercept will be located at the origin as well. So we will go ahead and label that on our graph. We'll go ahead and approximate 12 to be here. OK, now that we've discussed all the possible information we can get from the original function, let's go ahead and move on to the first derivative. Now, again, the first derivative is given to us as Y is equal to 2/5X2 multiplied by 9 minus X. So, how, what can we get from the first derivative? Well, the information that we can get from the first derivative is going to be the intervals of increasing and decreasing. What we want to do is you want to go ahead and find the values of the critical points for X. In order to do this, we are going to take the first derivative and set it equal to 0. Now, similar to the X intercepts, we are going to set each factor of X equal to 0. If 2/5 X 2 is equal to 0, simplifying that will give us X's equal to 0. And 9 minus X equal to 0 will give us the value of X's equal to 9. So what this means is that we have 2 potential critical points of the graph, but now we want to determine if these critical points are going to be local minimums or local maximums. So, we will go ahead and create a number line and plot our potential critical points here. Next, what we are going to do is we are going to pick testing points on the intervals from negative infinity to 0, between 0 and 9, and from 9 to positive infinity. Testing points that we can pick are going to be -1, 1, and 10. What we are going to do is we are going to plug in these values into the derivative. If the value of the derivative is negative, then the function will be decreasing in that region, and if the value of the derivative is positive, then the function will be increasing. Plugging in -1 into the first derivative will give us a positive value. So we are going to be increasing within this region. Plugging in positive 1 into the derivative will also give us a positive value. So, we are going to be increasing within the region from 0 to 9. And plugging in 10 into the derivative will produce a negative value. So we will be decreasing. From 9 to infinity. Notice that when we pass over the value of 0, the function does not switch from increasing to decreasing. So 0 is not going to be any critical point or local maximum or local minimum. But when we pass through 9, we switch from increasing to decreasing, indicating that X equals to 9 is the location of a local maximum. If we plug in the value of 9 into the original function, we get F of 9 is approximately equal to 218.7. So what this means is that we have a local maximum located at 9,218.7. We will go ahead and label this on the graph now. We'll say it's roughly here. OK, so we just labeled our local maximum. Now what we are going to do is we are going to determine the concavity of the function. In order to determine the concavity, we will use the second derivative. The second derivative was given to us as Y is equal to 65 X multiplied by 6 minus X. In order to find the inflection points, we are going to set the 2nd derivative equal to 0. Again, setting both factors of X equal to 0, we are going to get X's equal to 0, and 6 minus X equal to 0 will give us X's equal to 6. What these are are potential inflection points of the graph. Now, just like the first derivative, we are going to plot these inflection points on a number line. And pick testing points to plug into the second derivative. Testing points we are going to pick is -1, 1, and 7. If the second derivative with the testing point plugged in is positive, we will be concave up within that region, and if it is negative, we will be concave down. Plugging in -1 into the second derivative. Will give us a negative value. So we are going to be concave down within the region of negative infinity to zero. If we plug in the value of 1 into the second derivative, we will get a positive value. So we are going to be concave up within the region of 0 to 6. And plugging in 7 into the second derivative will give us a negative value. So we are going to be concave down in the region from 6 to to infinity. Since we are switching concavity from negative when we pass through 0, that means that 0 is going to be an inflection point. And since we are switching concavity from positive to negative, passing through 66 is also an inflection point. So, let's go ahead and plug in these values into the original function to find the exact location of the inflection points. Now, we do know where 0 is located at, Si X is equal to 0, that means that the inflection point is going to be the origin. But if we plug in the value of 6 into the original function, we get an approximate value of 129.6. So, what this means is that we have an inflection point located at 6.129.6. Let's go ahead and label this on our function for the graph. 6, and we have 129, which is approximately located. Let's say halfway here. Now if we just connect the dots. This is going to be the shape of the given function. And that is the final graph of the original Y function. So I hope this video helps you in understanding on how to approach this problem, and I will go ahead and see you all in the next video.

Graphs and Graphing

Graph the curves in Exercises 33–42.

y = 𝓍³ (8―𝓍 )

Key Concepts

Polynomial Functions

Factoring and Roots

Graphing Techniques

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