Question Video: Graphing a Cubic Polynomial Using Derivatives Mathematics • Higher Education

Use derivatives to identify which of the following is the graph of the function 𝑓(π‘₯) = π‘₯Β³ + 2π‘₯Β² + 3? [A] Graph A [B] Graph B [C] Graph C [D] Graph D [E] Graph E

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Video Transcript

Use derivatives to identify which of the following is the graph of the function 𝑓 of π‘₯ is equal to π‘₯ cubed plus two π‘₯ squared plus three? Options (A), (B), (C), (D), and (E).

In this question, we’re given a function 𝑓 of π‘₯ and we’re asked to determine which of five given graphs represents the graph of this function. And the easiest way to answer this question is to eliminate options by using properties of the function. For example, we can determine the 𝑦-intercept of the graph of the function 𝑓 of π‘₯ by substituting π‘₯ is equal to zero into the function. If we do this, we get 𝑓 of zero is equal to zero cubed plus two times zero squared plus three, which we can evaluate is equal to three. Therefore, the 𝑦-intercept of the graph of this function must be at three. And we can see option (A) does not have a 𝑦-intercept of three. Option (C) does not have a 𝑦-intercept of three. Option (D) also does not have a 𝑦-intercept of three.

We can continue using this logic to determine which of the two given options is correct, option (B) or (E). One way of doing this is to note in option (B) there’s an π‘₯-intercept at negative one. This means that 𝑓 evaluated at negative one should be zero. And if we substitute negative one into the function 𝑓 of π‘₯, we get negative one cubed plus two times negative one squared plus three. And we can evaluate this; it’s equal to four. Therefore, π‘₯ is equal to negative one is not an π‘₯-intercept of the graph of 𝑓 of π‘₯. So option (B) cannot be correct. This is then enough to conclude that option (E) must be the correct sketch.

However, there’s a few problems with this method. First of all, it requires us to be given options so that we can use process of elimination. Secondly, the question specifically asks us to use derivatives, and we’ve not used derivatives to evaluate the graph of this function. And finally, it’s a very useful skill to be able to sketch graphs of functions. So instead, let’s try and answer this question by sketching the graph of 𝑓 of π‘₯. And we’ll do this by analyzing the given function 𝑓 of π‘₯. Let’s start by clearing the five given options and taking note we’ve already shown that the 𝑦-intercept of this graph should be at three.

The next thing we can notice is our function 𝑓 of π‘₯ is a polynomial. In particular, the highest degree of π‘₯ is three. So it’s a degree three polynomial, which is called a cubic. And this allows us to determine the domain and the range of our function. First, the domain of any polynomial is all real values of π‘₯. This means when we sketch our function, we’ll sketch it for all values of π‘₯. Second, we can recall our degree polynomials have opposite end behavior. This means the range of the cubic function will always be all real values. We could now try and determine the roots of the function 𝑓 of π‘₯. One way of doing this would be to substitute values of π‘₯ into the function and see which ones are close to zero.

For example, we can evaluate 𝑓 at negative three, negative two, negative one, and zero. We would find that 𝑓 of negative three is negative six, 𝑓 of negative two is three, 𝑓 of negative one is four, and 𝑓 of zero is three. We can see that the outputs of the function change sign between π‘₯ is equal to negative two and π‘₯ is equal to negative three. And we can conclude that this means there’s a root between π‘₯ is equal to negative two and π‘₯ is equal to negative three.

We could analyze the end behaviors of the function to determine that this is the only root. However, this is not the only way; we can also use derivatives. And since the question asked us to use derivatives, we’ll use this method. We’ll just take note that we’ve shown there’s a root, or π‘₯-intercept, of the graph between π‘₯ is equal to negative three and π‘₯ is equal to negative two.

To use derivatives to analyze this function, we need to find the first derivative of 𝑓 of π‘₯. We’ll do this by using the power rule for differentiation. We multiply each term by the exponent of π‘₯ then reduce the exponent of π‘₯ by one. This gives us that 𝑓 prime of π‘₯ is equal to three π‘₯ squared plus four π‘₯. And there are several different things we can use the first derivative of a function to determine about its graph. First, we recall the critical points of a function are the points where the derivative is not defined or where the derivative is equal to zero. And in the case of polynomials, the derivative always exists. So the only critical points will be when its first derivative is zero. We want to solve three π‘₯ squared plus four π‘₯ equals zero.

We can do this by taking out the shared factor of π‘₯ on the left-hand side of the equation. This gives us that π‘₯ times three π‘₯ plus four is equal to zero. And now we can solve this equation by solving each factor equal to zero. We get π‘₯ is equal to zero, or π‘₯ is equal to negative four over three. These are the π‘₯-coordinates of the two critical points of our function where its slope is zero. This means these will be turning points of the graph of the function. We can find the exact coordinates of the turning points by substituting these values into our function 𝑓 of π‘₯.

First, we’ve already evaluated 𝑓 at zero; it’s output was three. So one of our critical points will be the 𝑦-intercept zero, three. Next, if we evaluate 𝑓 at negative four over three, we would get 113 over 27, which its decimal expansion is 4.185 recurring. This is too specific to get exactly correct in our sketch. However, it is worth taking note of where the critical points lie, and it can be often useful to write these as decimal expansions, the point negative 1.3 recurring, 4.185 recurring.

And before we move on to the second derivative, we could also recall this is not the only thing we can use the first derivative to find. We could also determine the intervals of increase and decrease of the graph of the function by using the sign of its first derivative. And the easiest way to do this is to sketch the graph of three π‘₯ squared plus four π‘₯. We’ll do this by using its factored form.

It’s a quadratic with positive leading coefficient, and we have its two roots at zero and negative four over three. Its graph would look like the following. And now we can easily see when π‘₯ is between zero and negative four over three, the sign of the first derivative is negative. And when π‘₯ is less than negative four over three or greater than zero, the sign of the first derivative is positive.

Therefore, we’ve shown the graph of 𝑓 of π‘₯ will be increasing when π‘₯ is less than negative four over three or π‘₯ is greater than zero, and it will be decreasing when π‘₯ is between negative four over three and zero.

Let’s now consider the second derivative of our function. This will measure the rate of change of the first derivative. It measures the rate of change of the slope of the curve. We can use this to determine the convexity of the curve. We find 𝑓 double prime of π‘₯ by differentiating 𝑓 prime of π‘₯ with respect to π‘₯ term by term by using the power rule for differentiation. We get 𝑓 double prime of π‘₯ is equal to six π‘₯ plus four.

We can determine when the curve is convex downwards by determining when its second derivative is positive, in other words, when is the slope of the curve increasing. We need to solve six π‘₯ plus four is greater than zero. And we can solve this inequality by subtracting four from both sides and dividing through by six and simplifying. We get that π‘₯ must be greater than negative two over three.

Therefore, when π‘₯ is greater than negative two over three, the second derivative of 𝑓 of π‘₯ is positive; its slope is increasing. In other words, the curve is convex downwards. On this interval, all of the tangent lines will lie below the curve. We can do the same to determine where the curve is convex upward. We want to solve six π‘₯ plus four is less than zero, and we see this is when π‘₯ is less than negative two over three. So on this interval, the tangent lines will lie above the curve since the slope of the curve is increasing on this interval.

And there’s one final thing we can notice. The convexity of the curve has switched when π‘₯ is equal to negative two over three. And we call points on the curve where its convexity switches inflection points. So the curve has an inflection point when π‘₯ is equal to negative two over three. And we could determine the exact coordinates of this inflection point by substituting negative two-thirds into our function 𝑓 of π‘₯. However, this is not necessary.

There is one final thing we can analyze about our function without using derivatives. We could look at the end behavior of the curve. We can see that 𝑓 of π‘₯ is a cubic polynomial with positive leading coefficient. So as the values of π‘₯ approach ∞, the outputs will approach ∞. And as the values of π‘₯ approach negative ∞, the outputs will also approach negative ∞. And this agrees with our analysis from the first derivative.

We can now use all of this information to sketch our curve. First, we start with a pair of coordinate axes, and we can add the two critical points onto our curve. This also includes the 𝑦-intercept at zero, three. We know the slope of our function will be zero as it passes through these two points. We also know there’s no more turning points of the curve. And as π‘₯ approaches ∞, the outputs of our function approach ∞. Similarly, as the values of π‘₯ approach negative ∞, the outputs approach negative ∞.

We can also label a few more key pieces of information on this diagram. For example, we’ve shown that the only π‘₯-intercept of this curve will lie between negative two and negative three. We could also mark on the coordinates of the inflection point of this curve. However, as mentioned before, it’s not strictly necessary since we already know that this matches option (E). Therefore, by using derivatives, we were able to identify that the correct graph of the function 𝑓 of π‘₯ is equal to π‘₯ cubed plus two π‘₯ squared plus three was given in option (E).

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