In Exercises 9–66, graph the function using appropriate method...

Question

In Exercises 9–66, graph the function using appropriate methods from the graphing procedures presented just before Example 9, identifying the coordinates of any local extreme points and inflection points. Then find coordinates of absolute extreme points, if any.
46. y = cos(x) + √3 * sin(x), 0 ≤ x ≤ 2π

46. y = cos(x) + √3 * sin(x), 0 ≤ x ≤ 2π

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to graph F of X, which is equal to the square root 3 multiplied by cosine of X, plus sin of X on 0 is less than or equal to X, which is less than or equal to 2 pi. The 1st and 2nd roots are given, and we're given FX, which is equal to minus the square root of 3 multiplied by sin of x, plus cosine of X, and F of X, which is equal to minus the square of 3 multiplied by cosine of X minus sine of X. Below the problem we're get an empty graph on which to plot our function. Now, the first thing we need to do is determine the domain of our function F of X, and we're actually given the interval between 0 and 2 pi, so we just need to make sure that our function is defined on that interval. And if we look at our function F of X, we see that it is a summation of cosine functions and sine functions, both of which are defined for all real values of X. So that means our function of F of X is defined everywhere on the interval from 0 to 2 pi. So that means that the domain of our function. Is going to be the closed interval from 0 to 2 pi. So next we need to determine our extrema. And here we'll be looking for local extrema, and for that we'll need the critical points and recall that the critical points occur whenever our first derivative is equal to zero or undefined. If you look at F of X again, it is made up of sine and cosine functions, and those are defined everywhere. So that means that F of X is defined everywhere on our domain. So the only time that we're going to have critical points is when F of X is equal to 0. So we're going to set minus the square root of 3, multiplied by a sine of X. Plus cosine of X equal to 0 and solve for X. The first step is to subtract cosine of X from both sides. And we'll have minus the square root of 3 multiplied by sin of X is equal to minus cosine of X. Now to software X we'll divide both sides of the equation by cosine of X and we'll also divide both sides of the equation by minus the square 3. In doing so, we'll have the tangent of X is equal to 1 divided by the square root of 3. Now recall that tangent of X is equal to 1 divided by the square root of 3 at 2 points on the interval between 0 and 2 pi, and those occur when X is equal to pi divided by 6, and when X is equal to. 7 pi divided by 6. So now we need to check to make sure that we have local extrema at these two critical points, and for this we'll use the second derivative test. So we're going to evaluate F at each of these points. So the first point we'll evaluate is X equal to pi divided by 6. So we'll calculate F of pi divided by 6, and this is going to be equal to minus the square root of 3, multiplied by the cosine of pi divided by 6. Minus the sign of pi divided by 6, and when we evaluate this expression, this is going to be equal to -2. Now, since our second derivative here is negative at this critical point, that means that we have a local maximum at this point. Next, we'll look at our other critical point with X equal to 7 divided by 6. So we'll calculate F of 7 pi divided by 6, and this is going to be equal to minus the square root of 3, multiplied by the cosine of 7 pi divided by 6 minus the sign of 7 pi divided by 6. And when we evaluate this expression, this is going to be equal to 2. So here a second derivative is positive at this critical point. So that means that we have a local minimum at this point. So we do have local extrema at both of our critical points. So now we just need to calculate the value of our function F of X at each of these points. So first, we'll look at F of pi divided by 6, and this is going to be equal to the square root of 3, multiplied by the cosine of pi divided by 6. Plus sign of pi divided by 6. And when we evaluate this expression, this is going to be equal to 2. And for the other point, we'll calculate F of 7 pi divided by 6, and this is going to be equal to the square root of 3, multiplied by the cosine of 7 pi divided by 6, plus sine of 7 pi divided by 6. And evaluating this expression, this is equal to minus 2. So we have a local maximum at the point of divided by 6.2, and a local minimum at the point of 7 divided by 6 minus 2. Next, we need to look for inflection points, and recall that inflection points occur whenever our second derivative F of X is equal to 0. So, in order to find those points, we're going to set minus the square root of 3. Multiplied by cosine of X minus sine of X. Equal to 0 and solve for x. So the first thing we'll do is add sin of x to both sides of the equation. So we'll have minus the square root of 3, multiplied by cosine of X is equal to sin of x, and dividing both sides of the equation by cosine of X, we'll have tangent of X is equal to minus the square root of 3. Now we just need to find the values of X, at which tangent of X is equal to minus the square of 3 on the interval between 0 and 2 pi. And recall that there are two values at which this occurs. This occurs when X is equal to. 2 pi divided by 3, and when X is equal to 5 pi divided by 3. So these are just possible inflection points. We need to make sure that the concavity is changing at these two points. And so we're going to look at the intervals around these two points and see if the concavity is changing. So the first interval that we're going to look at is going to be between 0 and 2 pi divided by 3. And on this interval, we just need to pick a point and evaluate our second derivative. So we're gonna choose pi divided by 6, and we've already actually calculated F of pi divided by 6. And we found that this was equal to -2. And here, our second derivative is negative, and recall that if the second derivative is negative on an interval, that means that our function F of X is going to be concave down on that interval. So our function FX is concave down on the interval between 0 and 2 pi divided by 3. The next interval that we're going to look at is going to go from 2 pi divided by 3 to 5 pi. Divided by 3. And again, we'll choose a point in this interval, and we're going to choose 7 pi divided by 6. And so we'll calculate F of 7 pi divided by 6, and we actually calculated this earlier and this was equal to 2. And since the second derivative here is positive, that means that our function is going to be concave up on this interval. Now the last interval that we want to look at is going to be going from 5 pi divided by 3 to 2 pi. And here we'll need to choose a point on the symbol, and so we'll choose X equal to 11 pi divided by 6. And so we'll calculate F of 11 pi divided by 6. And this is going to be equal to. Minus the square root of 3, multiplied by cosine. Of 11 pi divided by 6. Minus the sign of 11 pi divided by 6. And evaluating this expression, this is going to be equal to -1. So again, here, our second derivative is negative over this interval, so that means that our function is going to be concave down over this interval. So we do see that our Concavity is changing at X equal to 2 divided by 3, and it's also changing at X equal to 5 divided by 3. So we do have inflection points at each of these locations. So now we just need to calculate the value of our function at each of these locations. So we'll look at F of 2 pi divided by 3. And this is going to be equal to the square root of 3, multiplied by the cosine. Of 2 pi divided by 3. Plus the sign. Of 2 pi divided by 3. And when we evaluate this expression, this is going to be equal to 0. And we'll also need to look at F of 5 pi divided by 3, and this is going to be equal to the square root of 3, multiplied by the cosine of 5 pi divided by 3. Plus the sign of 5 pi divided by 3, and when we evaluate this expression, this is going to be 0 as well. So our reflection points occur at the point of 2 divided by 3.0 and the point of 5 pi divided by 3.0. These also happen to be RX intercepts. And now, the final quantity that we need to calculate are the values of our function at our end points. So we'll need to calculate. F of 0, and that's going to be equal to the square root of 3, multiplied by the cosine. Of 0. Plus the sign of 0. And if I win this expression, this is equal to the square root of 3. Which is approximately equal to 1.73. Likewise, we need to calculate F of 2 pi. That's going to be equal to the square root of 3, multiplied by the cosine of 2 pi. Plus the sign of 2 i. This is also equal to the square root of 3, which is approximately equal to 1.73. So now we have all the information that we need to create our graph. So we're going to start off by plotting our end points at 0.1.73. And at 2 1.73. Next, we'll plot our maxim that occurs at pi divided by 6.2. And we'll plot our local minimum that occurs at 7 pi divided by 6 minus 2. And then finally, we'll place our um inflection points on a graph, and that occurs at 2 pi divided by 3.0 and 5 pi divided by 3.0. And so now all we need to do is connect these points with a smooth curve. And so we'll begin at X equal to 0 and we'll increase until we reach our local maximum at X equal to pi divided by 6. And We then begin to decrease, crossing the x-axis at our inflection point, in which case we're changing from concave down to concave up. We will continue to decrease until we reach the local minimum. Then our function will begin to increase again. Until we reach the next inflection point, which is also where we cross the x axis at 5 pi divided by 3, and then our function will continue to increase until we reach the other end point. At To pie And so the graph that we have drawn on the screen here is going to be the answer to this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

Key Concepts

Graphing Trigonometric Functions

Local Extreme Points

Inflection Points

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