Video: Finding the Graph of a Cubic Function by Finding Its Turning Points Using Differentiation

Which of the following is the graph of the function 𝑓(π‘₯) = βˆ’π‘₯Β³ + 6π‘₯Β² βˆ’ 9π‘₯ + 1?

05:16

Video Transcript

Which of the following is the graph of 𝑓 of π‘₯ equals negative π‘₯ cubed plus six π‘₯ squared minus nine π‘₯ plus one?

We begin by observing that our function 𝑓 of π‘₯ is a cubic polynomial. And importantly, we know that its leading coefficient, that is, the coefficient of π‘₯ cubed, is negative. We know that when this is the case, then our cubic graph will have the general shape on the right of the two diagrams. We note that as π‘₯ tends to negative ∞, the function will tend to positive ∞. And as π‘₯ tends to positive ∞, the function will tend to negative ∞.

So looking at the five possible graphs we’ve been given, we can eliminate three straight away. Graphs A, C, and E are all graphs of cubic polynomials. But they correspond to cubics which have a positive leading coefficient. We’re therefore left with only graphs B and D.

Next, we could consider the constant term in our cubic polynomial, which is positive one, as this gives the value of the 𝑦-intercept of the graph. However, looking at graphs B and D, they both have a 𝑦-intercept of positive one. So this doesn’t help in determining which is the correct graph.

What we can do though is use differentiation to find the critical points of our function 𝑓 of π‘₯. Because we can see that the critical points on the graphs of B and D are different. So if we can find the critical points of 𝑓 of π‘₯, we can determine which is the correct graph. We recall then that the critical points for a function occur when its first derivative, which in this case will be 𝑓 prime of π‘₯, is equal to zero or does not exist.

Now our function 𝑓 of π‘₯ is a cubic polynomial. And so its derivative will be defined for all values of π‘₯. So we don’t need to be concerned about possible points where its derivative may not exist. Let’s find an expression for 𝑓 prime of π‘₯ using the power rule of differentiation.

We recall that, to differentiate each term, we multiply by the original exponent and then reduce the exponent by one. So the derivative of negative π‘₯ cubed is negative three multiplied by π‘₯ squared. The derivative of positive six π‘₯ squared is positive six multiplied by two π‘₯. The derivative of negative nine π‘₯ is negative nine, which we can see if we think of negative nine π‘₯ as negative nine π‘₯ to the first power. And the derivative of any constant is simply zero, which we can see if we think of one as one multiplied by π‘₯ to the power of zero. So our expression for 𝑓 prime of π‘₯ is negative three π‘₯ squared plus 12π‘₯ minus nine.

Next, we need to set this expression for 𝑓 prime of π‘₯ equal to zero. And we see that we have a quadratic equation in π‘₯. We can simplify somewhat by dividing through by negative three, giving the quadratic π‘₯ squared minus four π‘₯ plus three is equal to zero. And this quadratic can be solved by factoring.

As the coefficient of π‘₯ squared is one, the first term in each of our factors will be π‘₯. And we’re then looking for two numbers which sum to the coefficient of π‘₯ β€” that’s negative four β€” and which multiply to the constant term of positive three. Those numbers are negative three and negative one. So the factored form of our quadratic is π‘₯ minus three multiplied by π‘₯ minus one is equal to zero.

We then recall that if the product of two factors is zero, at least one of the factors themselves must be zero. So we set each set of parentheses in turn equal to zero and solve the resulting linear equation. π‘₯ minus three equals zero leads to π‘₯ equals three. And π‘₯ minus one equals zero leads to π‘₯ equals one. This tells us that the π‘₯-coordinates of the critical points of our function are one and three.

Now let’s look back at the two graphs, looking at graph B first of all. We can see that this does indeed have turning points at π‘₯-values of one and three. It has a local minimum when π‘₯ equals one and a local maximum when π‘₯ equals three. Looking at graph D, however, we can see that the π‘₯-coordinates of these critical points are negative one and negative three. There’s a local maximum when π‘₯ equals negative one and a local minimum when π‘₯ equals negative three. This tells us that it must be graph B which represents 𝑓 of π‘₯.

If we want to be even more sure, we could evaluate the function itself at each of our critical points. Substituting π‘₯ equals one into the equation of 𝑓 of π‘₯, we find that 𝑓 of one is equal to negative three, which is indeed the correct value for the function at the critical point on graph B. In the same way, if we were to substitute π‘₯ equals three, we would find that 𝑓 of three is equal to one, which again confirms that we’ve chosen the correct graph.

So by first considering the general shape of this cubic graph, which has a negative leading coefficient. And then using differentiation to find the π‘₯-coordinate of its critical points, we’ve found that the graph which represents the function 𝑓 of π‘₯ is graph B.

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