Video: Identifying the Graph of a Trigonometric Function

Which of the following is the graph of the function 𝑓(π‘₯) = 2π‘₯ sin 2πœ‹π‘₯ on the interval [βˆ’1, 1]? [A] Graph A [B] Graph B [C] Graph C [D] Graph D [E] Graph E

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Video Transcript

Which of the following is the graph of the function 𝑓 of π‘₯ equals two π‘₯ times sin two πœ‹π‘₯ on the closed interval negative one to one?

We’ve been given five graphs which could represent our function two π‘₯ sin of two πœ‹π‘₯ over the closed interval negative one to one. There are a few steps we can take to decide which of the graphs we’re actually interested in. To begin with, we’re simply going to set our function equal to zero. This will tell us the location of any π‘₯-intercepts. In other words, two π‘₯ sin of two πœ‹π‘₯ is equal to zero.

Well, this function itself is actually the product of two individual functions. It’s the product of two π‘₯ and sin of two πœ‹π‘₯. And so for the product of two functions to be equal to zero, either one or other of those functions must themselves be equal to zero. So we say either two π‘₯ is equal to zero or sin of two πœ‹π‘₯ is equal to zero.

Let’s divide our first equation by two so that π‘₯ is equal to zero. And we can solve our second equation algebraically, or we can simply consider the shape of the sine curve. We know the graph of 𝑦 is equal to sin of π‘₯ looks a little something like this. We see it’s equal to zero at negative πœ‹, zero, πœ‹, and two πœ‹.

Now we know that since a sine function is periodic, these values of π‘₯ will repeat. We’ll have negative two πœ‹, negative three πœ‹, and so on. For now, we’ll just consider the fact that two πœ‹π‘₯ could be equal to negative two πœ‹, negative πœ‹, zero, πœ‹, and two πœ‹. We divide through by two πœ‹. And we find π‘₯ is equal to negative one, negative one-half, zero, one-half, and one.

Now, of course, we were only interested in values of π‘₯ on the closed interval from negative one to one. So we don’t need to consider any other points. And we’ve found that there are five π‘₯-intercepts for our function. They are negative one, negative one-half, zero, one-half, and one.

Now, in fact, given our graphs, this would actually be enough to answer the question. But we’ll consider some alternative options. We can consider the value of the 𝑦-intercept by setting π‘₯ equal to zero and finding 𝑓 of zero. Well, 𝑓 of zero is zero. And so we now know the location of the 𝑦-intercept of our graph. But what about the turning points or the critical points of our graph?

We recall that the critical points of a function are points where the first derivative, 𝑓 prime of π‘₯, is either equal to zero or does not exist. So let’s find the first derivative of our function. Since the function is the product of two differentiable functions, we’re going to use the product rule. We’ll let 𝑒 be equal to two π‘₯ and 𝑣 be equal to sin of two πœ‹π‘₯. Then d𝑒 by dπ‘₯ is two. We also know that if we differentiate sin of π‘₯, we get cos of π‘₯. So the derivative of 𝑣 with respect to π‘₯ is two πœ‹ times cos of two πœ‹π‘₯.

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‘₯. In this case, that’s two sin of two πœ‹π‘₯ plus four πœ‹π‘₯ times cos of two πœ‹π‘₯. Now, this is actually defined for all π‘₯. So let’s set it equal to zero and solve for π‘₯.

This is not a particularly easy equation to solve. So we’ll use the functionality on our calculator. And when we do, we get π‘₯ is equal to positive and negative 0.781 and so on and positive and negative 0.322. There’s one other solution, and that’s π‘₯ is equal to zero. So this tells us the π‘₯-coordinate of any critical points.

We see that there’s a total of five critical points or turning points. And so we’re looking for a graph which has π‘₯-intercepts at negative one, negative one-half, zero, one-half, and one. A 𝑦-intercept at zero. And five critical points at positive and negative 0.7819, π‘₯ equals positive and negative 0.322, and π‘₯ equals zero. In fact, the only graph which satisfies this criteria is (A). So we can say that the graph of the function 𝑓 of π‘₯ is equal to two π‘₯ sin of two πœ‹π‘₯ is (A).

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