Question Video: Using Derivatives to Determine the Graph of a Cubic Polynomial | Nagwa Question Video: Using Derivatives to Determine the Graph of a Cubic Polynomial | Nagwa

# Question Video: Using Derivatives to Determine the Graph of a Cubic Polynomial Mathematics • Third Year of Secondary School

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Using derivatives, which of the following is the graph of π(π₯) = 3π₯ β π₯Β³?

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### Video Transcript

Using derivatives, which of the following is the graph of π of π₯ is equal to three π₯ minus π₯ cubed?

And in this question, we are given five options, which means we can eliminate options to determine the correct graph. However, for due diligence, itβs more useful to work out how we would sketch the given graph and work out all of the information we can determine by using derivatives and the given function π of π₯ is equal to three π₯ minus π₯ cubed for its graph.

So, instead, letβs use the given function π of π₯ is three π₯ minus π₯ cubed and the properties of the derivatives to determine the shape and information about its graph. This will allow us to sketch the graph. And we can then at the end of the video look at the five given options to determine which one best represents our graph.

Letβs start by looking at the given function π of π₯. Itβs three π₯ minus π₯ cubed. We can notice this is a cubic polynomial. And we can recall polynomials are defined for all real values of π₯. So the domain of this function is all real values. And because this is a cubic polynomial, its range will also be all real values. And this is because the polynomial is of odd degree. If it was even degree, this would not be true. We would need to determine the range. This is just a more formal way of saying that the graph will have one of the general cubic shapes.

Letβs now determine more information about the graph by using the function. We can substitute π₯ is equal to zero into our function π of π₯ to determine the coordinates of the π¦-intercept. π of zero is equal to three times zero minus zero cubed, which evaluates to give us zero. So the origin is the π¦-intercept of this curve. We can determine the π₯-intercepts by solving the equation three π₯ minus π₯ cubed is equal to zero. Both terms on the left-hand side of this equation share a factor of π₯.

So we can take out the shared factor of π₯ to get π₯ times three minus π₯ squared. We could factor the left-hand side of the equation further by using the difference between squares. However, this is not necessary. We can just solve each factor equal to zero. We get π₯ is equal to zero or π₯ is equal to positive or negative root three. And since we know cubic polynomials can have at most three π₯-intercepts, this is all of the π₯-intercepts.

Therefore, weβve shown that the graph of this function will intersect the π₯-axis at the points zero, zero; root three, zero; and negative root three, zero. And itβs worth pointing out this also includes the π¦-intercept, the point zero, zero.

Letβs now clear some space. We could add these three points onto a pair of coordinate axes and then try and sketch a cubic function which passes through these three points. However, this is not yet enough information to sketch an accurate diagram of our curve. So, instead, letβs carry on looking for more properties that our curve must have.

Another thing we can consider is, does the graph of this function have any symmetries, for example, vertical or horizontal lines of symmetry or a rotational symmetry? And itβs easy to see that no cubic function will ever have a horizontal or vertical line of symmetry. And this is because the two end behaviors of any cubic curves are opposite. And if we reflect this curve through the π₯-axis or through the π¦-axis, we will switch the end behaviors. So no cubic curve will have a horizontal or vertical line of symmetry.

However, we can check if the function is odd, which would mean it has rotational symmetry around the point zero, zero of order 180 degrees. To do this, we want to check if π evaluated at negative π₯ is equal to negative π of π₯. And we can do this by substituting negative π₯ into the function π. We get three times negative π₯ minus negative π₯ cubed. And we can then evaluate and rearrange this equation. Evaluating, we get negative three π₯ plus π₯ cubed. We can then take out a factor of negative one to get negative one times three π₯ minus π₯ cubed. And we can see this is just equal to negative π of π₯.

Therefore, weβve shown that the function π of π₯ is odd, which means the graph of this function has rotational symmetry about zero, zero 180 degrees clockwise or counterclockwise. So we can clear some space and add this onto our list of properties.

Weβre almost ready to start analyzing the derivatives of our function. However, thereβs one more thing we can look at. We can consider the end behavior of the graph of this function and if the function has any horizontal or vertical asymptotes. First, we know that cubic functions wonβt have any horizontal or vertical asymptotes. Next, we can recall the end behavior of any polynomial is determined by the sign of its leading coefficient and its degree. In this case, weβre dealing with a cubic polynomial and the leading coefficient is negative one. So, as the values of π₯ increase without bound, the outputs of our function are going to decrease without bound. As π₯ approaches β, the outputs are going to approach negative β. Similarly, as π₯ approaches negative β, the outputs are going to approach positive β.

This is now enough information to know the general shape that our function will have, since it has three π₯-intercepts and itβs a negative leading coefficient cubic. However, we want to sketch an accurate diagram. This means weβre going to need to know information about the critical points of the function and things like inflection points. And to analyze things like inflection points and critical points, weβre going to need to use derivatives.

To do this, letβs start by finding the first derivative of π of π₯ with respect to π₯. To do this, weβre going to need to use the power rule for differentiation. For each term, we multiply by the exponent of π₯ and then reduce this exponent by one. In this case, π prime of π₯ is equal to three minus three π₯ squared. We can recall that the critical points of any function are where the derivative is equal to zero or the derivative does not exist. Of course, for polynomials, the derivative always exists. So we just need to find where π prime of π₯ is equal to zero.

So we want to solve the equation three minus three π₯ squared is equal to zero. We can do this by dividing through by three to get one minus π₯ squared is equal to zero. And then we can just solve this equation. We add π₯ squared to both sides and take the square root. We get π₯ is positive or negative one. So this function has critical points when π₯ is positive and negative one. Since π prime of π₯ is equal to zero at these two points, the slopes of the tangent lines to the graph at these two points will be zero. These will be turning points.

We can also find the exact coordinates of these critical points by substituting π₯ is equal to positive and negative one into our function π of π₯. And we can just evaluate these. π evaluated at one is three times one minus one cubed, which is equal to two. And π evaluated at negative one is three times negative one minus negative one cubed, which is equal to negative two. And it is worth noting this agrees with what we found before. Our function π of π₯ is odd. So π evaluated at one and π evaluated at negative one should have opposite signs.

So letβs clear some space and make note that the only critical points of our curve should be at the points one, two and negative one, negative two. We could now move on to the second derivative. However, this is not the only piece of information we can determine from the first derivative of our function. We can also recall the graph of our function will be increasing when π prime of π₯ is positive and it will be decreasing when π prime of π₯ is negative. And we can use the critical points of our function to help us determine the intervals of increase and decrease.

Letβs start with the intervals of increase. We want to determine when three minus three π₯ squared is greater than zero. Since three π₯ squared is always positive, we can see this will happen when three π₯ squared is less than three. One way of seeing this is to solve the inequality. Three minus three π₯ squared is greater than zero. So we add three π₯ squared to both sides. So three is greater than three π₯ squared. We can then divide through by three. We get that one must be greater than π₯ squared. And then thereβs several different ways we can solve this inequality. We could do this graphically. However, itβs easier to do this by inspection. If one is greater than π₯ squared, then π₯ squared must be less than one, which means that π₯ must be between negative and positive one.

We can follow the same process to determine the intervals of decrease. We want to find when π prime of π₯ is less than zero. If we do this, we find that this will be when π₯ is greater than one or when π₯ is less than negative one. So we can add this information on to our list.

Now, letβs move on to the second derivative. We differentiate π prime of π₯ with respect to π₯ by using the power rule for differentiation. π double prime of π₯ is equal to negative six π₯. We can then recall the second derivative of a function measures the rate of change of its slope. In particular, the sign of the second derivative tells us the convexity or concavity of the curve.

Letβs start by determining when the second derivative is positive. That means that negative six π₯ will be greater than zero. We can just divide both sides of the inequality through by negative six. Well, remember, we need to switch the direction of the inequality. We get π₯ is less than zero. Since the second derivative is positive on this interval, we can conclude the graph will be convex downwards. The slope of the function is increasing.

We can do the same to determine when the second derivative is negative. We need to solve negative six π₯ is less than zero, so we divide through by negative six. We get that π₯ must be greater than zero. So, when π₯ is greater than zero, the slope of the function is decreasing. Its graph is convex upward. So we can add this information on to our list. The curve is convex downward when π₯ is less than zero and convex upward when π₯ is greater than zero. And of course this allows us to notice one final thing. The convexity of the curve is switching when π₯ is equal to zero. This means π₯ is equal to zero is a point of inflection of the curve. The convexity changes.

So letβs clear some space and now use this information to sketch a graph of this function. We can start by adding in the three π₯-intercepts at negative root three, zero, and root three. This also includes the π¦-intercept. And this also includes our points of inflection. Now, letβs add in the two critical points. These are the points with coordinates negative one, negative two and one, two. These will be the turning points of our function, where the slope of the tangent lines are zero.

Now, all we need to do is add in a cubic curve with a negative leading coefficient as shown. This then gives us the following sketch. We can now check which of the five given options matches this graph. Letβs start with option (A), where we can mark a few of the useful coordinates. We can see in option (A) the critical points seem to be at the points negative one, two and one, negative two. This disagrees with what we found about our function. The critical points should be at one, two and negative one, negative two. This is enough to conclude this is not the correct option. We can also note that the end behavior of the graph is incorrect. And in a similar way, we can notice the function is not increasing and decreasing or convex upward or convex downward on the correct intervals.

So letβs now move on to option (B), where weβve added in a few key pieces of information. We can see the end behavior of the graph is correct. And we can also see the three π₯-intercepts are also correct. However, if we consider the π¦-coordinates of the two critical points of our curve, we can see that these are between one and negative one. However, we showed that the critical points need to be at the points one, two and negative one, negative two. So option (B) cannot be the correct option. The critical points are too close to the π₯-axis.

So letβs instead move on to option (C), where once again weβve added some key information onto this diagram. We can note that the π₯-intercepts do appear to be at negative root three, zero, and root three. However, there are many things wrong with this diagram. For example, the end behavior is incorrect and once again the critical points are not at the correct position. We can see that the π¦-coordinates of the two critical points are between negative one and one. However, weβve shown they need to be at two and negative two. So, because the end behavior is incorrect and the critical points are not in the correct position, we can determine this is not the correct graph. And of course we can also notice the increasing and decreasing intervals are incorrect and the convexity is also incorrect.

So letβs now move on to option (D), where once again weβve added some key information onto our graph. We can read off the π₯-intercepts of this graph. Theyβre at negative one, zero, and one. And once again, we can also consider the π¦-coordinates of the two critical points of this curve. The π¦-coordinates of these two critical points are between one and negative one. So, once again, this canβt be the graph of our function. It has the wrong π₯-intercepts, and it has the wrong critical points.

So that just leaves us with option (E), which looks like the following. If we look at the π₯-intercepts of this graph, we can see that theyβre approximately at negative root three, zero, and root three. Similarly, we can consider the two turning points. From the diagram, they appear to be at the points one, two and negative one, negative two. The general shape appears to be a cubic function, and we can also see that the end behavior of the graph agrees with what we found. As π₯ approaches negative β, the outputs approach β. And as π₯ approaches β, the outputs approach negative β.

Finally, we could also consider where the function is increasing, decreasing, and its convexity by using the given diagram. However, this is not necessary. Instead, we can just conclude that option (E) gives the correct graph of the function π of π₯ is equal to three π₯ minus π₯ cubed.

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