Graphing functions Use the guidelines of this section to make ...

Question

Graphing functions Use the guidelines of this section to make a complete graph of f.

f(x) = x³ - 6x² - 135x

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to create a complete graph for the function F of X is equal to X cubed, minus 9X2 minus 108 X. So, in order to create a graph of function, we'll need several properties of a function. The first thing we'll look at is the domain. And so, if I look at our function F of X, it is a polynomial, and polynomials are defined over all real values. That means our domain is going to be the open interval from minus infinity to infinity. Now, the next quantity that we can look at is symmetry. So, to check for even or odd symmetry, I'm going to calculate F of minus X. This is going to be equal to the quantity of minus X in quantity cubed, minus 9 multiplied by the quantity of minus X in quantity squared, minus 108 multiplied by minus X. So this is going to be equal to minus X cubed. Minus 9 X squared. Plus 108 X. And this is not equal to F of X, nor is it equal to minus F of X. So that means that we have neither even nor odd symmetries. We can't really use symmetry to help us graph our function. Now, the next quantity that we want to look for are our critical points. For this we'll need the first derivative, so we'll calculate F of X. Which is equal to the derivative with respect to X. Of the quantity of X cubed minus 9 X2 minus 108 X. So using our constant multiple and power rule to take this derivative, this is going to be equal to. 3 X 2 -18 X. -108. Now we call to determine critical points, we need to determine the X values which our first derivative is either undefined. Here we have a polynomial, so it's defined everywhere, but the second condition is when our first derivative is equal to zero. So we will set. 3 X2 minus 18 X. -108 equal to 0. Now, to solve this for X, I'll first divide both sides by 3. And so we'll have X2 minus 6 X minus 36 is equal to 0. And here we can solve this using our quadratic formula. In doing so, we'll have X is equal to, and here we call your quadratic formula, we'll have minus 6, plus or minus the square root of the quantity of minus 6 squared, minus 4 multiplied by 1, multiplied by minus 36 in quantity. And all that is divided by the quantity of 2 multiplied by 1. So we can simplify this expression, so this is going to be equal to. 6 Plus or minus, and in the square root, what's inside the square root sign, I'll have my 6 squad, which is 36, so I can factor out a 36 out of both of those terms. And underneath the square root sign, and the square root of 36 to 6, so we'll have 6 multiplied by the square root. Of the quantity of 1 + 4 in quantity that is divided by 2. So simplifying further, this is going to be equal to. 3 Plus or minus, 3 multiplied by the square root of 5. So here we have two answers. So we have X equal to 3 plus 3 multiplied by the square root of 5. And this is approximately equal to 9.71. And the other solution Is X is equal to 3 minus 3 multiplied by the square root of 5, and this is approximately equal to -3.71. So those are our two critical points. So now, we're going to use these critical points to determine the intervals over which our function F of X is increasing or decreasing. So for this, we're going to make a small little table. So in the first row, we'll have the sign of F of X. And in the second row, we'll have the behavior of F of X, whether it's increasing or decreasing. So the first interval that we're gonna look at is from minus infinity. 2 X equal to minus 3.71. And so we're going to do is choose a value of X in this interval, so we'll choose X equal to -5. So calculating F of minus 5, we get 57, which is a positive quantity. And since our first derivative here is positive over this envolt, that means our function F of X is increasing, so we'll label that with an I in our table. Now the next interval that I want to check is the interval between -3.71 and 9.71. So here again, we'll choose a value of X, and we'll choose X equal to 0 and calculate F of 0. And F 0 is equal to -108, which is a negative quantity. We'll put a minus sign there in the first row, and here we call that if our first derivative is negative, that means our function F of X is decreasing. So we'll label that with a D in our table for the second row. And now the last interval that I want to check is the interval going from 9.71 to infinity, and here we'll choose a value for X and we'll choose X equal to 10. So, calculating F 10, we get a value of 12, which is a positive quantity. That means our function F of X is increasing over this interval. So we've identified the intervals over which our function is increasing and decreasing, and based on these intervals, we see that we have a local maximum at X equal to -3.71, and a local minimum at 9.71. So the next quantity that we need to look at is our concavity as well as our inflection points. So here we'll calculate F of X. In order to calculate this quantity, so this will be the derivative with respect to X. Of the quantity of 3 X squad minus 18 X. -108. In quantity. So again, applying our constant multiple and power rules take this derivative, this is going to be equal to. 6 X -18. And here when we take the derivative with respect to X of -108, since it's a constant, its derivative is going to be equal to 0. So to find possible inflection points, we need to set our second derivative equal to 0 and solve for X. So we'll have 6 X minus 18 equal to 0. And solving this for X, we'll get 18 divided by 6, which is going to be X is equal to 3. So we need to check the intervals of concavity for X less than 3 and X greater than 3. So for X, less than 3. We're going to choose a value of X. And calculate our second derivative. So we're gonna choose X equal to 0. That means that F of 0. It's going to be equal to. -18, which is less than 0 since it's negative, and since our second derivative here is negative, that means that our function F of X is going to be concave down. Recall that if our second ro is negative, our function F of X is concave down. Now, the other info that we need to check is X, greater than 3. And here we'll choose a value for X to calculate our second derivative, so we'll choose X equal to 4. What have F double prime of 4? That's going to be equal to 6, which is greater than 0. And so, here our second derivative is positive, so that means that our function is concave up on this interval. So that means we do have an inflection point at X equal to 3, since our cod cavity is changing from down to up. So now we have all the information that we need to actually create a graph. So one hour empty graph. We're going to actually label a couple of points. So, the first point that we're going to look at is when X is equal to -3.71. So when we calculate F of -3.71, We get approximately 225.7. So we're gonna plot that point. On our graph Now, the next point we're going to look at is X equal to 0. And so when we Calculate F of 0, we get 0, so that's actually our Y intercept. The next point that we're going to look at. is when X is equal to 3. So this is going to be our inflection point. So calculating F3, that is equal to -378, so we'll plot that point on the graph. The next point that we're gonna look at is when X is equal to 9.71. So calculating F of 9.71, we get approximately equal to minus 980. 1.7. So we'll mark that point on the graph. And now we have enough information to actually create our graph. And so we're gonna use the fact that Going from minus infinity to -3.71, our function is increasing, and it's also concave down on this interval. We're gonna draw a curve that is. Increasing, but concave down. And it's going to reach a maximum. At X equal to -3.71. Then our function is going to decrease, going through the origin. And it's going to reach an inflection point at X equal to 3. And it's going to continue decreasing, but at this point it's gonna be concave down. Until it reaches a minimum at X equal to 9.71. And then our function is going to increase after that. And so we'll just continue to draw an increasing function until It reaches off the graph. And the result that we have is going to be the answer to this problem. So the graph that we've drawn here is the correct answer. So, I hope that this explanation has been useful, and I'll see you in the next video.

Graphing functions Use the guidelines of this section to make a complete graph of f.f(x) = x³ - 6x² - 135x

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Graphing functions Use the guidelines of this section to make a complete graph of f.

f(x) = x³ - 6x² - 135x