49–54. {Use of Tech} Graphing with technology Make a complete ...

Question

49–54. {Use of Tech} Graphing with technology Make a complete graph of the following functions. A graphing utility is useful in locating intercepts, local extreme values, and inflection points.

ƒ(x) = 3x⁴ + 4x³ - 12x²

Accepted Answer

Hi everyone. Let's take a look at this practice problem. This problem says to graph the function F of X, which is equal to x of the 4th, minus 13 x cubed, plus 36X2d and we're also given FX, which is equal to 4 x cubed minus 39 x 2 plus 72 X, and we're given FX, which is equal to 12x2 minus 78 X plus 72. Below the problem we're given an empty graph on which to plot our function. So in order to graph our function F of X, we need to analyze this function. So the first thing we need to do is determine its domain. And if we look at a function F of X, we see that it's a polynomial, and recall that polynomials are defined for all values of X. So that means our domain is gonna be the open interval from minus infinity to infinity. Next, we need to check for symmetries. So we'll check for symmetries by calculating F of minus X. And this is going to be equal to the quantity of minus X in quantity to the 4th, minus 13 multiplied by the quantity of minus X in quantity cubed, plus 36 multiplied by the quantity of minus X in quantity squared. And so this is going to be equal to x 4th, plus 13 x cubed. Plus 36 x 2. And so this is not equal to. F of X And it is not equal to minus F of X. And so, therefore, we don't have even nor odd symmetries, so we can't really use any symmetries to help us draw or graph. The next quantity that we want to look at are our critical points, and recall that our critical points occur for x values whenever our first derivative is either equal to zero or undefined. If we look at our first derivative F of X, we see that it's a polynomial, so it's defined for all values of X. So the only possible critical points are going to be due to the first root of being equal to 0. So we're going to set 4 X cubed minus 39 x 2 + 72 X. Equal to 0. So we can factor out an X out of each of those terms. So we'll have X multiplied by the quantity of 4 X squad minus 39 X plus 72 in quantity, and that's going to be equal to 0. So we can set each one of these factors equal to 0, and then solve for X. So we'll have for the first factor X equal to 0, so that's one critical point, and then we'll set 4X2 minus 39 X plus 72 equal to 0. And so we're going to need to use the quadratic formula in order to solve this quadratic equation. So we call for the quadratic formula that we have X is equal to. In this case, we'll have the quantity of 39 plus or minus the square root of the quantity of 39 squared minus 4 multiplied by 4 multiplied by 72 in quantity, and all that is divided by the quantity of 2 multiplied by 4. So this will give us two solutions, and we'll evaluate this expression using our calculator. And so when we use the minus sign, and the quadratic formula, we'll get X is equal to. 2.47, and we use the plus sign, we'll get X is equal to 7.28. And here we just rounded to two decimal places. So we actually have 3 critical points here. And so now we're going to use these critical points to determine the intervals over which our function is increasing or decreasing. And to do this, we're going to use a table. So our table is going to consist of two rows. The first row is going to be the sign of FX, and the second row is going to be the behavior of FX, whether it's increasing or decreasing. And so we're going to look across several intervals. So our first interval is going to go from minus infinity to 0. And to check the sign for FX in this interval, we're going to choose the value of X and then calculate FX. So if we choose an X value of -1, and calculate F-1, we'll see that its value is equal to -115. And so that is a negative value, so we'll put in minus sign on the first row. And since our first derivative is negative, recall that that means that our function F of X is decreasing on this interval. So we'll label that with a D in our table. The next interval that we're going to look at is going to go from 0 to 2.47. And again, we'll choose a value for X to calculate our first derive, so we'll choose a value of X equal to 1, and when we calculate F of 1, we get a value of 37, which is positive. And since our first derivative is positive on this interval, that means our function F of X is increasing. So, the next interval that we're going to look at is going to be going from 2.47 to 7.28. And again, we'll choose a value in this interval to calculate the first derivative, so we'll choose X equal to 3. And that means that F 3 is going to be equal to -27, so that is a negative value. And since our first derivative is negative, that means our function F of X is going to be decreasing. And then the final interval that we want to check is going to be going from 7.28 to infinity. And again, choosing a value, we'll choose X equal to 8, and calculating F 8, we get a value of 128, which is a positive value, and that means that our function F of X is increasing on this interval. So here we've identified all the intervals over which our function F of X is decreasing or increasing, and we see that since our function is changing from decreasing to increasing at X equal to 0 and X equal to 7.28, that those are going to be local minima and we're going to have a local maximum at X equal to 2.47 since our function is changing from increasing to decreasing. So the next quantity that we need to calculate. our inflection points, and for this we'll need our second derivative. And recall that inflection points occur whenever our second derivative is equal to 0. So, we're gonna set the quantity. Of 12 X squad. Minus 78 X. Plus 72 equal to 0, since that is our second derivative. And we're gonna solve this quadratic equation for X. So again, we'll have to use the quadratic formula, so we'll have X. is equal to the quantity of 78 plus or minus the square root of the quantity of 78 squared minus 4 multiplied by 12, multiplied by 72 in quantity. And all that is divided by. The quantity of 2 multiplied by 12 in quantity. So again, we'll evaluate this expression using our calculators. And when we take the minus sign, we'll get X is equal to approximately 1.11. And when we take the positive sign, X is going to be equal to 5.39. Again, here we just round it to two decimal places. So this, we're going to use these um possible values as inflection points in order to determine the intervals of concavity. And again, we're going to use a table to determine this. So, again, our table will have two rows. The first row is going to be the sign of our second row of F of X, and the second row is going to be the behavior of F of X. So, the first interval that we'll look at is going from minus infinity. 2 1.11. And here we're going to have to choose a value for X in order to calculate the second derivative in this interval. So, we'll choose X equal to 0, and when we calculate F of 0, we get a value of 72, so that is a positive value. So we'll put a plus sign in the first row of our table. And recall that if the second derivative is positive, that means that our function F of X is going to be concave up. So we'll label that with a U in the second row of our table. The next area that we want to look at is going from 1.11 to 5.39. And again, we'll choose a value in this interval, so we'll choose X equal to 3, and we'll calculate F of 3, that's going to be equal to -36. So that is a negative value. And recall that if our second derivative is negative, that means that our function F of X is concave down, so we'll label that with a D. And then the final interval that we want to look at is going from 5.39 to infinity. Again, choosing the value in this interval will choose X equal to 6, and F of 6 is equal to 36, and so that is a positive value. And so that means that our function F of X is concave up on this interval. Now, the last quantities that we want to look at are our intercepts. So, to find our Y intercept, we'll set X equal to 0. So, we'll look at F of 0, and that's going to be equal to 0. Next, we need to find our X intercepts, and for this we'll set our function F of X equal to 0. So we're gonna set X the 4th, minus 13 X cubed. Plus 36 x squared. equal to 0, and we need to solve this for X. So we can factor out an X2d out of each of those terms. So we'll have x2 multiplied by the quantity of X2 minus 13 X plus 36, and quantity is equal to 0, and we can actually factor that quadratic. And so we'll have X2d multiplied by the quantity of X minus 4 in quantity, multiplied by the quantity of X minus 9. In quantity, and that has to be equal to 0. So to find our X intercepts, we'll set each factor equal to 0 and solve for X. So we'll have X2 equal to 0, and so that means that X is equal to 0, which we already found as our Y intercept as well since it goes through the origin. The next factor is going to be X minus 4, equal to 0, and so that means that X is equal to 4, and the final factor will have X minus 9 equal to 0, so we'll have X equal to 9. And so, now we have all the information that we need in order to create our graph. So we're gonna start off by plotting our intercepts, our inflection points, and our critical points. So, the first point that we'll plot is our Y intercept, which is at the origin, so that's gonna be 0.0. The next point is going to be Our inflection point at 1.11, and we calculate F of 1.11, we get approximately 28.1. So we'll plot that point on the graph. The next point is going to be a critical point at X equal to 2.47. And When we calculate F of 2.47, we get approximately 61. So we'll plot that point as well on the graph. The next point is going to be our X intercept at X equal to 4. Then we'll have our next critical point. I'm sorry, our next inflection point at X equal to 5.39. And when we calculate F of 5.39, we get approximately minus 145.8. We'll plot that point in our graph as well. And then, the next point that we're going to look at is going to be our Next critical point, at X equal to 7. 28 And we calculate F of 7.28, we get approximately minus 299. We'll plot that point on our graph. The next point that we'll look at is going to be RX intercept at X equal to 9. So now what we need to do is use the fact that When X is equal to plus or minus infinity, our function. F of X approaches infinity in both cases, since our highest power in our function of FF X is X the 4th, that's an even function. So whenever I square infinity or raise infinity to the 4th power, I get a positive quantity. And so, we'll have our function coming in from Infinity. And we call that our function is decreasing 4 X. Less than 0, and so we'll come down until we reach the origin. And that's going to be a local minimum. Then we're going to begin increasing until we hit the inflection point. And we'll continue increasing until we reach the local maximum. And then we'll begin decreasing, passing through our X intercept, and then to our next inflection point, and we'll continue decreasing until we reach our local next local minimum, and then our function will increase. Passing through the X intercept. And continuing on to infinity. So the graph that we have plotted on the screen is going to be the answer to this problem. So I hope that this explanation has been useful and I'll see you in the next video.

49–54. {Use of Tech} Graphing with technology Make a complete graph of the following functions. A graphing utility is useful in locating intercepts, local extreme values, and inflection points.

ƒ(x) = 3x⁴ + 4x³ - 12x²

Key Concepts

Graphing Functions

Intercepts

Local Extrema and Inflection Points

Watch next