Identifying Extrema

In Exerc...

Question

Identifying Extrema

In Exercises 53–60:

a. Find the local extrema of each function on the given interval, and say where they occur.

f(x) = √3cos x + sin x, 0 ≤ x ≤ 2π

Accepted Answer

Hi, everyone. Let's take a look at this practice problem. This problem says to find the local extrema of the function G of X, which is equal to sin of x minus the square root of 3, multiplied by cosine of X on the interval 0 is less than or equal to X, which is less than or equal to 2 pi. So we need to find the local extrema of our function G of X on our given interval between 0 and 2 pi. So, in order to find the local extrema, I first need to find the critical points, and to find the critical points, I need to calculate G of X. So G of X is going to be equal to the derivative with respect to X. Of the quantity of sin of X. Minus the square root of 3 multiplied by cosine of X in quantity. So taking this derivative, this is going to be equal to. When I take the derivative with respect to X of sin of X, I get cosine of X. Then we'll have minus the square root of 3 is a constant, so it'll move through the derivative, so we'll have the square root of 3, and that gets multiplied by the derivative with respect to X of cosine of X, and that derivative is equal to minus sine of X. So simplifying this expression, this is going to be equal to cosine of X, plus the square root of 3, multiplied by sine of X. So now we need to find the critical points and recall that the critical points occur whenever our first derivative is undefined or equal to zero. So if we look at our function G of X, we know that it's made up of sine and cosine functions, and both of those functions are defined everywhere. So that means that G of X is also defined everywhere. So the only time we're going to have critical points is when our first derivative is equal to 0. So we're going to set cosine of X plus the square root of 3 multiplied by sin of X. Equal to 0 and solve for X. The first step is to subtract cosine of X from both sides of the equation. So I'll have the square root of 3, multiplied by sine of X is equal to minus cosine of X. So now I need to divide both sides of this equation by the square root of 3, and divide both sides of the equation by cosine of X. Doing so, I will get the tangent of X equal to minus 1 divided by the square root of 3. Now recall that your tentative function is equal to -1 divided by the square root 3 whenever its argument is equal to minus pi divided by 6 plus an integer multiple of pi. So that means that X is going to be equal to minus pi divided by 6, plus n multiplied by pi, wheren here is an integer. So we need to find other values of X that fall within our interval between 0 and 2 pi. So, we're going to look at different cases of N. So we're gonna start off with N equal to 0, and when N is equal to 0, X is equal to minus pi divided by 6, which is a negative number which falls outside of our interval, since the lower bound is equal to 0. The next case we'll look at is when N is equal to 1, and when N is equal to 1, X is equal to 5 pi divided by 6. And this does fall within our info, so that is one of our critical points. The next case to look at is when N is equal to 2, and when N is equal to 2, X is going to be equal to 11 pi divided by 6, which is also between 0 and 2 pi. The next case we'll look at is when N is equal to 3, and when N is equal to 3, X is going to be equal to 17 pi divided by 6, but that is actually greater than 2 pi, which is our upper bound for our interval. So, these are going to be the only two critical points that we need to analyze, since they're the only two that fall within our interval. So, we need to determine whether these critical points are local minima or local maxima. And to do this, we're gonna look at this, we're going to use the second derivative test. So we need to calculate G of X. That's going to be equal to the derivative with respect to X. Of the quantity of cosine of X. Plus the square root of 3 multiplied by sine of X in quantity. So taking this derivative, this is going to be equal to. Minus sine of x. Plus the square root of 3 multiplied by cosine of X. And so what we need to do now is evaluate the second derivative at each of our critical points. So we need to look at G of 5 pi divided by 6. And this is going to be equal to minus sign of 5 pi divided by 6. Plus square root of 3, multiplied by cosine of 5 pi divided by 6. And when we evaluate this expression, this is going to be equal to minus 2. And here since our second derivative is negative at this critical point, that means that we're going to have a local maximum. So we call that if the second derivative is negative, that means we have a local maximum. So we have a local maximum at X equal to 5 pi divided by 6. Next, we need to evaluate G of 11 pi divided by 6. And this is going to be equal to minus sign of 11 pi divided by 6. Plus the square root of 3, multiplied by the cosine of 11 pi divided by 6. And evaluating this expression, this is going to be equal to 2. So here we see our second derivative is positive at this critical point. So that means that we have a local minimum at this point. So we have a local minimum at X equal to 11 pi divided by 6. So, each of our critical points do correspond to a local extrema. So now we need to evaluate our function G at these two points. And so, we need to calculate G of 5 pi divided by 6, and this is going to be equal to sin of 5 pi divided by 6. Minus the square root of 3, multiplied by cosine of 5 pi divided by 6, and when we evaluate this expression, this is going to be equal to 2. And we also need to evaluate G of 11 pi divided by 6 for the critical point, and this is going to be equal to sine of 11 pi divided by 6. Minus the square root of 3, multiplied by the cosine. Of 11 pi divided by 6. And evaluating this question, this is going to be equal to -2. So for a final answer, we're going to have a Local minimum At X equal to 11 pi divided by 6, with G of 11 pi divided by 6 equal to minus 2. And we have a local maximum. When x is equal to 5 divided by 6 with G of 5 pi divided by 6 equal to 2, so those are the answers that this problem was looking for. So I hope that this explanation has been useful, and I'll see you in the next video.

Identifying Extrema

In Exercises 53–60:

a. Find the local extrema of each function on the given interval, and say where they occur.

f(x) = √3cos x + sin x, 0 ≤ x ≤ 2π

Key Concepts

Local Extrema

First Derivative Test

Trigonometric Functions

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