Video: Graphing Using Derivatives of Given Functions

Consider the function 𝑓(π‘₯) = (π‘₯ βˆ’ 1)Β²(π‘₯ + 2). Find 𝑓′(π‘₯). Find and classify the critical points of 𝑓. Find the intervals of increase and decrease for 𝑓. Find lim_(π‘₯ β†’ ∞) 𝑓(π‘₯). Which of the following is the graph of 𝑓? [A] Graph A [B] Graph B [C] Graph C [D] Graph D [E] Graph E

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Video Transcript

Consider the function 𝑓 of π‘₯ equals π‘₯ minus one squared times π‘₯ plus two. Find 𝑓 prime of π‘₯. Find and classify the critical points of 𝑓. Find the intervals of increase and decrease for 𝑓. Find the limit as π‘₯ approaches infinity of 𝑓 of π‘₯.

And then there’s one more part of this question, which asks us to identify the correct graph for our function. So we’ll look at our options when we’ve completed the first four parts. So how are we going to evaluate 𝑓 prime of π‘₯?

That’s the first derivative of our function with respect to π‘₯. Well, taking a look at our function, there are a number of ways we can do this. We could distribute our parentheses and differentiate term by term. Alternatively, notice that we have a product of two functions, one of which is a composite function. So we could use the product rule alongside the general power rule.

The product rule says that the derivative of the product of two differentiable functions 𝑒 and 𝑣 is 𝑒 times d𝑣 by dπ‘₯ plus 𝑣 times d𝑒 by dπ‘₯. So let’s let 𝑒 be equal to π‘₯ minus one all squared and 𝑣 be equal to π‘₯ plus two. We’ll need to use the general power rule to find the derivative of π‘₯ minus one all squared. This is a special case of the chain rule. And it says we can multiply the entire function by the exponent, reduce the exponent by one, and then multiply all of this by the derivative of the inner function.

Well, the derivative of π‘₯ minus one is just one. So d𝑒 by dπ‘₯ is two times π‘₯ minus one to the power of two minus one times one. And that simplifies to two times π‘₯ minus one. d𝑣 by dπ‘₯ is much more straightforward. It’s simply one. So we substitute what we have into the product rule. It’s 𝑒 times d𝑣 by dπ‘₯ β€” that’s π‘₯ minus one all squared times one β€” plus 𝑣 times d𝑒 by dπ‘₯. Distributing our parentheses, and the first derivative of our function simplifies really nicely to three π‘₯ squared minus three.

In the second part of this question, we need to find and classify the critical points of our function. Remember, these occur when our first derivative is either equal to zero or does not exist. Well, we actually know that polynomial functions are all differentiable. So we’re just going to consider the case for which our first derivative is equal to zero. That is, three π‘₯ squared minus three is equal to zero.

Let’s solve for π‘₯. We divide through by one. And then we use the difference of two squares to factor π‘₯ squared minus one. It’s π‘₯ plus one times π‘₯ minus one. Now for the product of these two terms to be equal to zero, either π‘₯ plus one has to be equal to zero or π‘₯ minus one has to be equal to zero. And solving each equation for π‘₯, we obtain π‘₯ must be either negative one or π‘₯ equals one. So we have critical points where π‘₯ is either equal to positive or negative one.

Our next job is to classify these. We can perform the first derivative test and evaluate the first derivative just before and just after each critical point. Let’s add a table. We know that the first derivative is equal to zero when π‘₯ is equal to negative one or positive one. The first derivative evaluated at π‘₯ equals negative two is three times negative two squared minus three, which is nine. The first derivative evaluated when π‘₯ is equal to zero is three times zero squared minus three, which is negative three. And the first derivative evaluated at π‘₯ equals two is also nine. The function is increasing before π‘₯ equals negative one and decreasing after. So the critical point π‘₯ equals negative one is a local maximum. The opposite is true when π‘₯ equals one. So that must be our local minimum.

In the third part of this question, we’re looking to find the intervals of increase and decrease for our function. Let’s clear some space. Now this might seem awfully similar to what we just did. But at that point, we were simply considering the nature of the function at specific points. We now want to know over what intervals the function is increasing or decreasing. So we’re going to work out when the first derivative is less than zero, decreasing, or greater than zero, increasing.

We know the graph of 𝑦 equals three π‘₯ squared minus three looks a little something like this. We can see that the graph of the first derivative is greater than zero here and here. In other words, when π‘₯ is less than negative one or greater than one. The first derivative is less than zero here. That’s when π‘₯ is greater than negative one or less than one. So the intervals of increase are the open intervals from negative infinity to negative one and one to infinity. And the interval of decrease is the open interval from negative one to one.

And finally, we need to find the limit as π‘₯ approaches infinity of the function. We can do this by direct substitution. We see that as π‘₯ approaches infinity, the function itself also grows larger and larger. So the function also approaches infinity. Let’s now combine everything we’ve done to identify the graph of our function.

Which of the following is the graph of 𝑓? If we were to distribute the parentheses of our function, we’d see we have a cubic graph with a positive leading coefficient. In other words, the coefficient of π‘₯ cubed is positive. That tells us the shape of the graph will be a little something like this. We know it has critical points π‘₯ equals negative one and one and a maximum and minimum in those locations, respectively.

We can work out the value of our function at these points by substituting negative one and one in. And when we do, we see that 𝑓 of negative one is four and 𝑓 of one is zero. So we know the graph has a maximum at negative one, four and a minimum at one, zero. We also saw that it had intervals of increase over the open interval negative infinity to negative one and one to infinity and decrease over the open interval negative one to one. The only graph that satisfies all of these conditions is A. In fact, we could have very quickly spotted they could not have been D or E as these are graphs of quadratic functions. C is a cubic graph which has a negative coefficient of π‘₯ cubed. So we really only had two options.

Notice that we didn’t worry about evaluating the limits of our functions to look for asymptotes. As we know, a cubic graph doesn’t have any.

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