24–34. Curve sketching Use the guidelines given in Section 4.4...

Question

24–34. Curve sketching Use the guidelines given in Section 4.4 to make a complete graph of the following functions on their domains or on the given interval. Use a graphing utility to check your work.

{Use of Tech} ƒ(x) = x (x -1)e⁻ˣ

{Use of Tech} ƒ(x) = x (x -1)e⁻ˣ

Accepted Answer

Hello. In this video, we are going to be drawing the graph of the given function below. The function given to us is F of X equal to X multiplied by X minus 2 multiplied by E to the power of negative X. Now, in order to start this problem, let's go ahead and check the domain of the given function. Now, the domain or the function is in the form of a polynomial, but specifically a product of polynomials and exponential function. Now, we know that for the factors of X and X minus 2, those domains are going to be all real numbers, but what about the exponential value E to the negative X? Well, E negative X can be rewritten as 1 divided by E to the power of X, but no matter what value of X you plug in to the exponential, this exponential value is never going to be zero. So what this means is that the domain for the exponential is all real numbers as well. And with that being said, the domain of the overall function is going to be from negative infinity to positive infinity, representing all real numbers. Now, what we want to go ahead and do is we want to go ahead and find the X intercepts of this graph. Now, in order to find the X intercepts, we want to find the values of X that cause the function to be zero. In order to do this, we will take the function X multiplied by X minus 2, multiplied by E to the negative X, set it equal to 0, and solve for X. We will go ahead and use the 0 property and set each factor of X equal to 0. That will give us X's equal to 0, which is the first X intercept. That will also give us X minus 2 equal to 0, and E to the negative X equal to 0. Now, for the second equation, we can add 2 to both sides, and that will leave us with X's equal to 2, which is going to be our second X intercept. But what about the exponential function? Well, just like we stated, any value of X that we plug into the exponential is going to cause it to get closer to zero, but never be zero. So what this means is that there is no value of X that's going to cause an exponential to be zero, thus, we will disregard this factor, since there is no solution. So what this means is that we have two X intercepts. The first one is located at the origin, 0.0, and the second one is located at 2.0. Furthermore, since we have an X intercept at the origin, that also means that we have a Y intercept at the origin as well. So we managed to find the Y intercept while solving for the X intercept. Now that we found these pieces of information, let's go ahead and check the regions of the graph that is increasing and decreasing. In order to find these regions, we first need to find the critical points of the graph, and the critical points are obtained by taking the derivative of the function. So before we take the derivative of the function, let's go ahead and distribute X into X minus 2. That will allow us to rewrite the function to be F of X is equal to E to the negative X multiplied by the quantity X2 minus 2 X. Now what we want to do is you want to take the derivative by using the product rule. By using the product rule, we will get the derivative as E to the power of negative X multiplied by 2X minus 2, plus the quantity X2 minus 2X multiplied by negative E to the power of negative X. By factoring out an E to the power of negative X and combining like terms, the derivative is going to be E to the power of negative X, multiplied by negative X squared. Plus 4 x minus 2. Now what we want to do is we want to solve for the critical points of the graph. Now, in order to find the critical points, we will go ahead and find the values of X that cause the derivative to be zero. So, in order to find those values of X that cause the derivative to be zero, we will take our derivative E to the power of negative X, multiply by negative X2 + 4 X minus 2, and set it equal to 0. Just like the X and Y intercept, we will go ahead and take each factor of X and set it equal to 0 to solve. We will have E to the power of negative X equal to 0, and negative X2 + 4 X minus 2 equal to 0. Now, again, there is no value of X that you can plug into an exponential that will cause it to be zero, so we will go ahead and disregard the exponential since there is no solution. But for the second equation, let's go ahead and set the leading coefficient to be positive. In order to do this, we will divide the entire equation by -1, allowing us to rewrite the second equation as X2 minus 4 X + 2. Now, this is not factorable, but we can go ahead and solve for the values of X by using the quadratic equation. By using the quadratic equation, we will end up with two values of X. X is equal to 2, plus or minus the square root of 2. So these are going to be two potential critical points for the graph, but now we need to identify the critical points as either local maximums or local minimums, and in the process, we will also determine the regions of increasing and decreasing. So we will go ahead and create a number line. Now, on this number line, we are going to plot our two critical points. The first one being 2 minus the square root of 2, and the second one being 2 plus the square root of 2. Now what we want to do is you want to go ahead and take testing points around the neighborhoods of these two critical points. If we plug them into the derivative and the derivative is positive, then that region of the graph is increasing. And if we take a testing point from the neighborhood, plug it into the derivative, and the derivative is negative, then that region is decreasing. So, let's go ahead and pick some testing points. A testing point that we can pick to the left of 2 minus the square root of 2 is going to be 0. When we plug in 0 into the derivative, F 0 is going to give us -2. Since the value is negative, that tells us that the region of the graph to the left of 2 minus square root of 2 will be decreasing. Now, a value that we can pick between 2 minus square root of 2 and 2 plus the square root of 2 is going to be 1. When we plug in 1 into the first derivative, we get the value of E to the power of -1. An exponential is positive in this case. Therefore, the region of this graph will be increasing. Then we will go ahead and pick a testing point to the right of 2 plus square root of 2, and that testing point can be 4. Plugging in 4 into the derivative will give us the value of negative 2 multiplied by E raise to the power of -4. This is a negative number, so the graph is going to be decreasing within this region. So, if we take a look at the first derivative test, we can conclude that the regions of decreasing are going to be from negative infinity. To 2 Minus the square root of 2. And they are going to be from 2 plus the square root of 2 to positive infinity. And the regions of increasing are going to be from in between the region of 2 minus the square root of 2 to 2 plus the square root of 2. Furthermore, when we pass through the critical 0.2 minus the square root of 2, we are the graph is going to be switching from decreasing to increasing. What this implies is that we have a local minimum. At the value, X is equal to 2 minus the square root of 2. And when we pass through the critical 0.2 plus the square root of 2, the graph switches from increasing to decreasing. By the first derivative test, we have a local maximum located at X's equal to 2 plus the square root of 2. Now that we have the regions of the local maximum and local minimum, and the regions of increasing and decreasing, let's go ahead and find the concavity of the graph. Now, in order to find concavity of the graph, we need to use the second derivative. Now recall that the first derivative was F of X equal to the power of negative X multiplied by negative X2 + 4 X minus 2. We will take the derivative by using the the product rule again. F of X will be E to the power of negative X multiplied by -2 X + 4, plus negative X2 + 4 X minus 2 multiplied by negative E to the power of negative X. After factoring out E to the power of negative X and combining light terms, the second derivative is going to be E to the power of negative X multiplied by X2. Minus 6 X plus 6. Again, we will find the not critical points, the inflection points. By setting both factors of the second derivative equal to 0. Now, again, we know that if we set the exponential equal to 0, there will be no solution. So, let's just go ahead and find where X2 minus 6 X + 6 is equal to 0. So we will take that factor of X and set it equal to 0. Then, by the quadratic equation, we will get two values of X. X is equal to 3 plus or minus the square root of 3. Now we will go ahead and use the 2nd derivative test. We will plot these values on a number line. And again, similar to the first derivative test, we will take testing points around the neighborhoods of these inflection points and plug them into the second derivative. If we take the testing 0.1 and plug it into the second derivative, F1 is equal to the power of -1. This is a positive value. Therefore, the region of the graph is going to be concave up from negative infinity to 3 minus the square root of 3. Taking a testing point between the two inflection points, we can pick the value of 2. Plugging in 2 into the second derivative will give us the value of -2 to the power of -1. The region will be concave down because the second derivative is negative in this region. Then a testing point that we can pick to the right of 3 plus the square root of 3 is 5. If we plug in 5 into the second derivative, we get E to the power of -6, which is a positive value. So that means that the regions of concave up are going to be from negative infinity to 3 plus the square root of 3, and from 3 plus the square root of 3 to 5. And the region where the graph is concave down is is between the region of 3 minus square root of 3 and 3 + the square root of 3. So, just to recap, we found that the domain is all real numbers. We have X intercepts located at 0.0 and 2.0, with the Y intercept also being at the origin. The graph is going to have a local minimum at 2 minus the square root of 2, and a local maximum at 2 plus the square root of 2, and is decreasing. From negative infinity to 2 minus square root of 2 and from 2 plus the square root of 2 to infinity. The graph will also be increasing from 2 in the region of 2 minus square root of 2 to 2 plus the square root of 2. So let's go ahead and create an axis for the graph. Now, let's go ahead and label our X and Y intercept. That is going to be at the origin and at the point 2.0 for the X intercept. Now, let's go ahead and plot the local maximum and local minimum. Now, we know the X location of the local minimum, but we don't know the Y location. If we were to plug in the local minimum into the original function F of X, then that is going to give us an approximate value. Of 0.16. So we can say that the local minimum will be located roughly here. At 2 minus the square root of 2.0.16. Then, let's go ahead and plug in the local maximum into the original function. Plugging in the local maximum 2 plus the square root of 2 will give us an approximate value of. Negative sorry positive. Actually, I wrote this down wrong. This should be negative 0.16, and this is going to be positive, 0.46. That means that we can label the local maximum roughly around here. At 2 + square root of 2.0.46. Now we know that the graph is going to be decreasing from negative infinity to the local maximum, so The behavior of the graph is going to look roughly like this. Then it's going to be increasing. From the local maximum, local minimum to the local maximum, and then it will settle down roughly like this. So this is going to be the rough. This is going to be an estimate of the graph to the given equation, and this is going to be the solution to the problem. 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.

24–34. Curve sketching Use the guidelines given in Section 4.4 to make a complete graph of the following functions on their domains or on the given interval. Use a graphing utility to check your work.

{Use of Tech} ƒ(x) = x (x -1)e⁻ˣ

Key Concepts

Curve Sketching

Derivatives

Exponential Functions

Watch next

24–34. Curve sketching Use the guidelines given in Section 4.4 to make a complete graph of the following functions on their domains or on the given interval. Use a graphing utility to check your work.{Use of Tech} ƒ(x) = x (x -1)e⁻ˣ

Key Concepts

Curve Sketching

Derivatives

Exponential Functions

Watch next

24–34. Curve sketching Use the guidelines given in Section 4.4 to make a complete graph of the following functions on their domains or on the given interval. Use a graphing utility to check your work.

{Use of Tech} ƒ(x) = x (x -1)e⁻ˣ