Determine the end behavior of the following transcendental fun...

Question

Determine the end behavior of the following transcendental functions by analyzing appropriate limits. Then provide a simple sketch of the associated graph, showing asymptotes if they exist.

$f (x) = \sin x$

Accepted Answer

Hello there. Today we're going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Consider the transcendential function F of X equals of 2 X. What is the end behavior of this function by analyzing the appropriate limits? Sketch a graph of the function showing asymptotes if they exist. Awesome. So we're asked to solve for two separate answers for this particular prompt. So we're trying to figure out the end behavior of this function by analyzing the appropriate limits. That's our first answer. Now we're trying to solve for the end behavior. Our second answer is we're trying to sketch a graph of this specific function showing asymptotes if any exist. Awesome. So now that we know that we're trying to solve for the end behavior and create a sketch of this particular function for FFX, let us first start out by, you know, start to solve this problem by focusing our attention on analyzing the limits for the end behavior. So with that said, as we should recall a note for transcendential functions like sine functions, the limits as X approaches infinity or negative infinity, which is from positive infinity or negative infinity, do not settle into a single value due to their periodic nature. So, therefore, for F of X equals sin of 2 X, the limits as X approaches positive infinity, and as X approaches negative infinity, Do not exist. So the acronym that we will use is DNE does not exist in the conventual conventional sense because the function continues to oscillate between -1 and 1. So since we're dealing with a sign function, it's going to oscillate up and down. Between 1 and 1. Awesome. So let me say that again. So for transcendential functions, such as ours sign function that were given by the prom itself, the limits as X approaches positive infinity. Or let's use the word or instead, or negative infinity, it will do not settle into a single value. Or it does not settle to a single value specifically due to the periodic nature of the graph itself. So therefore for F of X equals sin of 2 X, the limits as X approaches positive infinity and X approaches negative infinity does not exist. So. That's why we can say and. So that's our first step right now is analyzing the end behavior. Our second step now. we need to identify the period of the function. So as we should recall and note, the general form of a sign function is, and let's write this in red, sign of be multiplied by X or BX, where the period is given by. 2 pi divided by B. So, for the case of our specific function, which is F of X is equal to sin of 2 X, in this case B will be equal to 2. So therefore, the period as a result will be 2 pi divided by 2, which will be equal to pi. So this means as a result, the function will complete one full cycle every pi or every value of pi units along the x axis. So once again, this is very important to note. So this will mean that our specific function of F of X will complete one cycle every pi units along the x axis. Awesome. So now our next step, since we're dealing with a sine graph, that will mean we have to determine the amplitude. So our next step now is we need to figure out the amplitude. And as we should recall and note, the amplitude of a sine function will be the coefficient in front of the sign value or the sign variable to be precise. Which will dictate its maximum and minimum values. So for the case of our specific function, which is F of X equals sin of of X, the coefficient will be 1, it's implied because we're not given a number multiplied outside of, so there's no number outside of the sign. So we will assume that it's 1. So this means that the function will oscillate between -1 and 1 on the Y axis. So let me make a little note of that off to the side here in blue. So the amplitude, which I'll call capital A, is going to be equal to 1 in this particular case. So now, our next step is we need to identify our asymptotes if there are any asymptotes. So, as we should recall and note, sign functions do not have vertical or horizontal asymptotes. So sine functions once again do not have vertical or horizontal asymptotes because they continue to oscillate indefinitely without approaching a specific Y value or moving infinitely far away from the X axis. So essentially when you think of like a a string like let's say. You like you strum a string on a guitar or like. You have a rubber band and you pull the rubber band down and it like starts to bounce up and down until until it goes to 0 or 10s to 0 would be the proper term in math, but it's gonna just keep oscillating until it goes to zero. So it's not gonna oscillate forever. It's gonna oscillate to the point where it just stops moving. That's what they mean, that it's not going to because they continue to oscillate indefinitely without approaching a specific Y value or moving infinitely far away from the X-axis. That's what it means by that. So now, moving right along here. So now that we know that there's no asymptotes, we can safely sketch the graph. So, when we sketch our graph, We can go ahead and note that we need to, in order to help us start officially sketching our graph, we need to start by making the X axis from, let's make a note of this in green, from negative to to to pi to show at least two full periods in order to be clear for clarification and also to make sure we have a nice size graph to look at and to plot. So we now, now that we know that we're making our X axis from negative 2 to 2 pi in order to show two full periods, we now need to mark our critical points where the function will intersect the X axis, and we need it to intersect at 0. I divided by 2. Hi. 3 pi divided by 2. And finally, at 2 pie. So we need to therefore now draw our sign curve starting from the origin point, which the origin point, and let's write this in red origin point will be at 0.0, where it's X. Y. So we need to draw our sine curve starting from the origin point, and it will reach a maximum at pi divided by 4, and it will cross the X axis again at pi divided by 2. Reaching a minimum value at 3 pi divided by 4, and this will continue in this specific pattern for at least 2 periods. Awesome. So putting what we said what we said in words into our graph, as you can see, we have our curve, the sine curve represented in red, and then we have our different coordinate plots represented by green dots and then the coordinates are written in green. So we have our origin 0.00 then we have our point, as we could say, our maximum at pi divided by 4. And then it crosses again at the on the x axis at pi divided by 2, and then it reaches a minimum at 3 pi divided by 4. Now as you can see, it continues this pattern going up and down, up and down. For two periods at least. So, with that said, Quickly recapping here. So in conclusion, our function of F of X equals sin of 2 X will oscillate indefinitely between -1 and 1 without approaching a specific limit as X approaches infinity or negative infinity, and it has a period of pi and has no asymptotes. And our graph once again is a sine wave that crosses the X axis at multiples of pi divided by 2, and it reaches its maximum and minimum at, and let's write this down. And also, before we write this down, let's create another graph. Really quick here and also right our limits. So really quick here, we need to note that it reaches its maximum and minimum at, and let's write these in blue at high divided by 4 plus K. Pi or K multiplied by pi. So once again, that is our maximum and our minimum will be 3 pi divided by 4 plus K multiplied by pi. Awesome. And where K in this case is an integer. So once again, this is our Max, and then this is our min. Which Let us make sure we write that nice like so we don't get confused about what's what. So this will be equal to our min. Awesome. So moving right along here. So once again, max is pi divided by 4, and our min is 3 pi divided by 4, and then now referring to our limits. So as the limit as X approaches positive infinity of sin of 2 X, it will does not exist for the end behavior. That's our end behavior does not exist. So let's put a box around that so it does not exist for our. I'll call it ED end or EB end behavior, so. EB and behavior. So then for our limit as X approaches negative infinity of sin of 2x it will does not exist. So that means that ultimately for our end behavior, our final answer is DNE does not exist in an hour. And finally, our second answer for our graph, we have our sign curve. And that's it. We've officially solved for this problem. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Determine the end behavior of the following transcendental functions by analyzing appropriate limits. Then provide a simple sketch of the associated graph, showing asymptotes if they exist. f(x)=sin⁡xf\left(x\right)=\sin x

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Determine the end behavior of the following transcendental functions by analyzing appropriate limits. Then provide a simple sketch of the associated graph, showing asymptotes if they exist.
$f (x) = \sin x$