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 - 147x + 286

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to draw a graph of the following function. If we're given the function F of X Z equal to X cubed minus 6 X2 + 9 X. The problem was us to create a graph of our function. So we're gonna need to determine a couple of properties of this function. The first thing we're going to look at is the domain of our function. And here for F of X, we have a polynomial. We call it polynomials are defined everywhere along the X axis. That means that our domain is going to be the open interval, going from minus infinity to infinity. Now, the next thing we want to check for is our symmetries, because symmetries will help us to actually draw our graph. So here we'll check for an even or odd symmetry by calculating F of minus X. In doing so, this is going to be equal to the quantity of minus X in quantity cubed. -6 multiplied by the quantity of minus X in quantity squared, plus 9 multiplied by minus X. And this is going to be equal to minus X cubed. Minus 6 X squad. Minus 9 X. Which is not equal to either F of X or minus F of X. So that means that our function here is neither even nor odd. So, we can't really use symmetries to help us draw our graph. The next thing we want to look at are the critical points for our function. And in order to calculate the critical points, we'll need our 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 6 X squared. Plus 9 X in quantity. So, here to take this derivative, we're going to use our constant multiple and power rules. And so this is going to be equal to. 3 x squared Minus 12 X. Plus 9. And so in order to find our critical points, we call that we need to determine the values of X, which our first derivative is undefined, which in this case, we have a polynomial, so our function here is our first derivative here is defined everywhere on the X axis, but the other condition is that when our first derivative is equal to zero. So we'll set 3 X squared. -12 X. Plus 9 is equal to 0. So we can actually divide both sides of this equation by 3, and do we say we'll get X squared. -4 X Plus 3 is equal to 0, and we can actually factor that quadratic. And so we'll have the quantity of X minus 1 in quantity, multiplied by the quantity of X minus 3 in quantity. That has to be equal to 0. So we're going to do is set each factor equal to 0 and solve for X. So we'll have X minus 1 is equal to 0, therefore X is equal to 1, and then we'll have X minus 3 is equal to 0, which means that X is equal to 3. So these here are our two critical points. So what we're going to do is use our critical points to determine the intervals over which our function is F of X is increasing or decreasing. So for this, we're going to actually create a table. So in the first row of the table, we'll have the sign of FX, whether it's positive or negative, and in the second row, we will have the behavior of F of X, whether it's increasing or decreasing. So the first interval that we're going to look at is going to be going from minus infinity to 1. And so, what we're going to do is choose a value for X in this interval, so we'll choose X equal to 0, and we'll calculate our first derivative value. So F of 0 is going to be equal to 9, which is a positive quantity. We will label that as a plus in our table. And we call that if the first derivative is positive over an interval, that means our function F of X is increasing. So we'll label that with an I in our table. So, our function F of X is increasing over this interval. Now, the next interval that we want to look at is the interval going from 1 to 3. So here we'll choose a point in this interval, so we'll choose X equal to 2, and we'll calculate F of 2. And that value turns out to be -3, which is a negative value, and if our first derivative is negative over an interval, recall that our function F of X is going to be decreasing over that interval. So we'll label that with a D in our table. And then the final interval that we need to look at is going from X equal to 3 to infinity. So again, we'll choose a point in this interval, so we'll choose X equal to 4. And so when we choose this, And calculate F of 4, we get a value of 9, which is positive. So that means that our function F of X is increasing over this interval. So now we've identified all the intervals over which our function F of X is increasing or decreasing. We also see that we have a local maximum at X equal to 1. Since in the interval just before X equal to 1, our function F of X is increasing, and in the interval just after it, our function is decreasing, which means we have a local maximum. Likewise, we see that at X equal to 3, we have a local minimum. Since in the interval just before X equal to 3, our function is decreasing, and just after X equal to 3, that interval is our function is increasing over that interval. So that means we have a local minimum. Now, the next quantity we want to look at are our inflection points. So for our inflection points, we'll need to calculate the second derivative. So we'll have F of X is equal to the derivative with respect to X of our first derivative, which is the quantity of 3 X squad minus 12 X. Plus 9 in quantity. So, here again, we're gonna need to use our constant multiple and power rules, as well as our constant rule for derivatives to take this derivative. So doing so, this is going to be equal to 6 X minus 12. And recall that to find inflection points, we need to set our second derivative equal to 0. We'll set 6 X minus 12 is equal to 0, and solving for X, we'll get X is equal to 2. So we're going to use this point to determine the intervals over which our function F of X is concave up and concave down. And so again, we'll make a small table here. In the first row we'll have the sign of the second row of F of X, and in the second row we'll have the behavior of our function F of X, whether it's concave up or concave down. And so, we actually have two intervals that we need to check. The first interval is going to go from minus infinity to 2. And again, we'll choose a point in this interval, so we'll choose X equal to 0. And when we calculate F 0, we get a value of -12, which is negative. We'll have a negative sign in the first row of our table. And recall that if the second derivative is negative over an interval, that means that our function over that interval is going to be concave down. So we'll label that with a D in our table to denote concave down. The next interval that we're going to look at is going from X equal to 2 to infinity. So here again, we'll need to choose a point in this interval. So we'll choose X equal to 3. And when we calculate F of 3, we get a value of 6, which is positive. And if we call it that if our second derivative is positive over an interval, that means that our function F of X is concave up on that interval. So we'll label that uh with a U in our table. So here we've defined all the intervals of which our function is either concave up or concave down, and we do see that we have an inflection point at X equal to 2, since our function is going from concave down to concave up at this point. So we now have enough information to begin to create our graph. So, we're going to use an empty graph. And we're gonna plot the points for our um critical points as well as our inflection points. So the first point that we want to calculate is X equal to 1. And so, when we calculate F of 1, We get a value of 4. And so we'll plot the point of 1.4. The next point that we want to look at is X equal to 2, and F of 2 has a value of 2, so we'll plot the point of 2.2. The next point we want to look at is when X is equal to 3, and F of 3 is equal to 0, so we'll plot the point of 3.0. And now we want to just add 2 more points just to help us um create the graph. So we'll look at X equal to 0. And when we calculate F of 0, we get a value of 0, so we'll plot a point at 0.0. And the next point that we want to look at is X equal to 4. And F of 4 is equal to 4, we'll plot the point of 4.4. Now we're going to use the fact that at the very beginning of our intervals, or X less than 1. That our function F of X is increasing, and it's also concave down. So we're going to draw a graph that is increasing over that interval, as well as concave down. And so we'll have a maximum occurring at X equal to 1. Our function is then going to decrease, and we're going to have an inflection point at X equal to 2. It will continue to decrease, but now our function is concave up. Until it reaches a local minimum at X equal to 3, and then our function is still going to be concave up, but increasing for the remaining. Um, X values. So, the graph that we have plotted on the screen is the answer to this question. 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 - 147x + 286

Key Concepts

Polynomial Functions

Critical Points and Extrema

End Behavior of Functions

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