Question Video: Identifying the Graph of a Polynomial Mathematics • Third Year of Secondary School

Video Transcript

Use derivatives to identify which of the following is the graph of the function 𝑓 of 𝑥 is equal to eight multiplied by negative 𝑥 squared plus 𝑥 all squared. Options (A), (B), (C), (D), and (E).

In this question, we need to identify which of five given graphs is the graph of the function 𝑓 of 𝑥 is equal to eight multiplied by negative 𝑥 squared plus 𝑥 all squared. And we’re told to do this by using derivatives. And it’s worth noting it’s possible to answer this question without using derivatives. One way of doing this would be to substitute values of 𝑥 into our function 𝑓 of 𝑥 and determine their outputs. We could then construct a function table to determine points which the graph of the function must pass through. And while this would work, the question specifically asks us to use derivatives. So let’s instead find an expression for 𝑓 prime of 𝑥. And to do this, let’s start by clearing some space and look at our function 𝑓 of 𝑥.

We can see that 𝑓 of 𝑥 is the composition of two functions. It’s the polynomial all squared. This means we have multiple different options to determine 𝑓 prime of 𝑥. One way we could do this is by using the chain rule. And this is a good way of finding 𝑓 prime of 𝑥. However, our outer function is only the square function, so we can also expand 𝑓 of 𝑥 by using the FOIL method. Either method would work, and it’s personal preference which one we want to use. In this video, we’re going to distribute the exponent over the parentheses. Doing this, we get that 𝑓 of 𝑥 is equal to eight multiplied by 𝑥 to the fourth power minus two 𝑥 cubed plus 𝑥 squared.

Now, we can differentiate 𝑓 of 𝑥 with respect to 𝑥 term by term by using the power rule for differentiation. We want to multiply by the exponent of 𝑥 and then reduce each exponent by one. Doing this gives us that 𝑓 prime of 𝑥 is equal to eight multiplied by four 𝑥 cubed minus six 𝑥 squared plus two 𝑥. And we can simplify this expression for 𝑓 prime of 𝑥 by noticing each of the three terms inside of the parentheses shares a factor of two and they also share a factor of 𝑥. This allows us to rewrite 𝑓 prime of 𝑥 as 16𝑥 multiplied by two 𝑥 squared minus three 𝑥 plus one.

Let’s now consider what the derivative tells us about the graph of the function. We can recall when 𝑓 prime of 𝑥 is equal to zero or does not exist, we say that the graph has a critical point. In particular, since in our case 𝑓 of 𝑥 is a polynomial, we know it’s defined for all real values of 𝑥. Similarly, its derivative is defined for all real values of 𝑥. So the critical points will only be when the derivative is equal to zero. Therefore, we can find all of the critical points of our function by solving 𝑓 prime of 𝑥 is equal to zero. We want to do this by factoring, so we’re going to need to factor the quadratic two 𝑥 squared minus three 𝑥 plus one. And there’s many different ways of doing this. We can note that this factors to give two 𝑥 minus one multiplied by 𝑥 minus one. And multiplying this by 16𝑥 gives us 𝑓 prime of 𝑥.

We can now solve 𝑓 prime of 𝑥 is equal to zero by setting each of the three factors equal to zero. Solving the first factor equal to zero gives us 𝑥 is equal to zero. Solving the second factor equal to zero gives us 𝑥 is equal to one-half. And solving the third factor equal to zero gives us 𝑥 is equal to one. Therefore, the graph of our function has three critical points: one when 𝑥 is zero, one when 𝑥 is one-half, and one when 𝑥 is equal to one. These will all be points on the graph where its derivative is zero, so the slope of the graph is zero, which means its tangent lines are horizontal at these points.

We can determine the exact coordinates of the critical points by substituting 𝑥 is zero, 𝑥 is one-half, and 𝑥 is equal to one into our function 𝑓 of 𝑥. If we do this and evaluate, we get that 𝑓 evaluated at zero is equal to zero; 𝑓 evaluated at one-half is equal to one-half; and 𝑓 evaluated at one is equal to zero. Therefore, the graph of our function has critical points at zero, zero; one-half, one-half; and one, zero. This is not the only thing we can use the derivative to determine. Remember, when 𝑓 prime of 𝑥 is positive, the graph of our function is increasing; it slopes upwards. However, when 𝑓 prime of 𝑥 is negative, the graph of our function slopes downwards; it’s decreasing.

Therefore, the sign of the first derivative will tell us when our function is increasing or decreasing. We can determine the sign of the first derivative by looking at our factored expression for 𝑓 prime of 𝑥. First, when all of the factors are positive, 𝑓 prime of 𝑥 will be positive. For the first factor to be positive, 𝑥 must be positive. For the second factor to be positive, 𝑥 must be greater than one-half. And for the third factor to be positive, 𝑥 needs to be greater than one. All three of these are satisfied when 𝑥 is greater than one, so 𝑓 prime of 𝑥 is positive when 𝑥 is greater than one. So the graph of our function is increasing when 𝑥 is greater than one.

Let’s now see what happens to the derivative of our function between two of our critical points, 𝑥 is one-half and 𝑥 is one. When 𝑥 is between one-half and one, 16 multiplied by 𝑥 is positive. Similarly, since 𝑥 is between one-half and one, two 𝑥 minus one will be positive because 𝑥 is greater than one-half. However, our third factor will be negative since 𝑥 is less than one. So we have a positive multiplied by a positive multiplied by a negative. Therefore, the first derivative of 𝑓 with respect to 𝑥 is negative on this interval. The graph of the function is decreasing when 𝑥 is greater than one-half and 𝑥 is less than one.

Let’s now follow the same reasoning for when 𝑥 is between the other two critical points, when 𝑥 is greater than zero and less than one-half. Since 𝑥 is greater than zero on this interval, we know that 𝑥 is going to be positive. So, 16 multiplied by 𝑥 is positive. However, on this interval, 𝑥 is less than one-half, so two 𝑥 minus one will be negative. Similarly, on this interval, our values of 𝑥 are less than one. This means 𝑥 minus one will be negative. So when 𝑥 is between zero and one-half, we have a positive multiplied by a negative multiplied by a negative. This gives a positive. So on this interval, the first derivative of 𝑓 is positive. Therefore, our function is increasing when 𝑥 is between zero and one-half.

There’s one final interval we want to consider: what happens when our values of 𝑥 are less than zero. When our values of 𝑥 are less than zero, 𝑥 is negative. So 16 multiplied by 𝑥 is negative. In exactly the same way, our values of 𝑥 are less than one-half and our values of 𝑥 are less than one. So two 𝑥 minus one is negative and 𝑥 minus one is also negative. So when our values of 𝑥 are less than zero, 𝑓 prime of 𝑥 is the product of three negative numbers, so it’s negative. This means the function is decreasing on this interval.

We could now analyze the second derivative to determine the intervals on which the graph of the function is convex upward and convex downward. And this is determined by the sign of the second derivative. We can also check for points where the convexity changes. These will be inflection points. However, as it turns out, it’s not necessary to analyze the second derivative to answer this question by using derivatives. So, although we could analyze the second derivative, we won’t in this case.

Let’s now look at the five given options in more detail. Let’s start with option (A). We can mark on any of the critical points of our function, the points where the slope is zero. These will be the points on the graph where the tangent lines are horizontal. We can see that there are three such points, and we can read off the coordinates of these three points: the point zero, zero; the point one-half, negative one-half; and the point one, zero. This is enough to conclude that this is not the graph of our function because, remember, we’ve already found the coordinates of all three critical points of the graph of 𝑦 is equal to 𝑓 of 𝑥. In particular, one of our critical points has coordinates one-half, one-half, not one-half, negative one-half. Therefore, this cannot be the graph of the function.

We can now follow the exact same process for option (B). Let’s start by marking its three critical points. These are the points where the slope of the graph is zero. We can see that there are three such points in option (B). However, we can see that two of the 𝑥-coordinates of these critical points are negative. These are not at the critical points we’ve found when 𝑥 is one-half and when 𝑥 is one, so option (B) is not correct. So let’s instead move on to option (C) and once again mark the three critical points. Once again, there are three critical points of this graph, and we can see one of these points has coordinates one-half, zero. And this is not the coordinates of one of the critical points we’ve found. So option (C) is not correct.

So let’s instead move on to option (D). Let’s once again mark the critical points of this graph. There are three critical points of this graph, and we can note one of them has coordinates negative one, zero. This does not agree with the three critical points we’ve found. So once again, we can eliminate this option. The answer cannot be option (D). This just leaves option (E). However, for due diligence, let’s check that the critical points are correct and the intervals of increase and decrease are also correct for this graph. First, we can see that the graph has three critical points: one at the origin, one at the point with coordinates one-half, one-half, and a final critical point at the point with coordinates one, zero. This agrees with the three critical points that we found earlier.

Next, let’s look at where the function is increasing and decreasing. We can see when our values of 𝑥 are greater than one, the graph of the function is sloping upwards; it’s increasing. Similarly, we can see when the values of 𝑥 are between zero and one-half, the graph of the function slopes upwards. It’s increasing. And this agrees with the two intervals of increase we found earlier. The function should be increasing when 𝑥 is greater than one and when 𝑥 is between zero and one-half.

In exactly the same way, we can determine the intervals where the graph of the function is decreasing. We can see when 𝑥 is less than one-half, the function is sloping downwards. It’s decreasing. Similarly, when the values of 𝑥 are between one-half and one, the graph is sloping downwards. It’s decreasing. And this agrees with the intervals of decrease we found earlier. The graph should be decreasing when 𝑥 is between one-half and one, and it should also be decreasing when 𝑥 is less than zero. Therefore, by using derivatives, we were able to identify which of five given graphs was the graph of the function 𝑓 of 𝑥 is equal to eight multiplied by negative 𝑥 squared plus 𝑥 all squared. It was given in option (E).