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) = sin x − cos 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 F of X, which is equal to sin of 2 X minus cosine of 2 X on the interval 0 is less than equal to X, which is less than or equal to pi. So we need to find the local extrema of our function F of X on the interval that was given to this problem. So in order to find the local extrema, we need to find our critical points. And to do that, we'll need our first derivative. So we'll need to calculate F of X, and that's going to be equal to the derivative with respect to X. Of quantity of sine of 2 X. Minus cosine of 2 X. In quantity. So taking this derivative, this is going to be equal to. And for the first term, we'll have the derivative with respect to X of sine of 2 X. Here I'm taking the derivative of a sine function, which will give me a cosine function. So we'll have the cosine of 2 X. We'll need to also apply your chain rule, so need to multiply this by the derivative with respect to X of 2 X, which is 2. Then we'll have minus for the second term, we'll have the derivative with respect to X of cosine of 2 X, and here we're taking the derivative of a cosine function. And when we take the derivative of a cosine function, we'll get a minus sine function. So we'll have minus sine. Of 2 X And then we'll need to also apply our chain rule here, so multiplying this by the derivative with respect to XF 2 X, which is 2. So simplifying our expression, this is going to be equal to 2 multiplied by the cosine of 2 X. Plus 2 multiplied by the sign of 2 X. So now we need to find our critical points and recall that our critical points occur whenever our first derivative is either undefined or equal to zero. And since sine and cosine function are continuous functions, and they're um defined for all values of X, the only time we're going to get critical points is when F of X is equal to 0. So we're gonna set 2 multiplied by cosine of 2 X. Plus 2 multiplied by the sine of 2 X. Equal to 0, and now we need to solve for X. So the first step is to subtract two multiplied by cosine of 2 X from both sides of the equation. So we'll get 2 multiplied by the sine of 2 X. And that's gonna be equal to -2, multiplied by cosine of 2 X. So now what we're going to do is divide both sides of the equation by cosine of 2 X and divide both sides of the equation by 2. In doing so, we'll get tangent of 2 X. is equal to -1. And we call that our tangent function is equal to -1 whenever our argument is equal to 3 pi divided by 4, plus an integer multiple of pi. So that means that 2 X. Which is our argument, it's going to be equal to 3 pi divided by 4, plus in pi. Wherein here is an integer. So we need to solve this for X by dividing both sides of the equation by 2. We'll get X is equal to. 3 pi divided by 8 plus N multiplied by pi divided by 2. And so now we need to determine the values of X that fall within our interval. So we're gonna start off with the case of N equal to 0. When N is equal to 0, we'll get X is equal to 3 pi divided by 8. So that's going to be our first critical point. When N is equal to 1, we evaluate the expression, we'll get X is equal to 7 pi divided by 8. So that's going to be our second critical point. And if you look at the case when N is equal to 2, We'll get a value for X of 11 pi divided by 8. However, that value is greater than pi, which is the upper bound of our interval. So these are going to be the only two critical points that we need to analyze. So now what we need to do is determine whether we have a local minimum or local maximum at each of these critical points. And to do that, we're going to use our second derivative test. So we're gonna need to calculate F of X. And that's going to be equal to the derivative with respect to X. Of the quantity of 2 multiplied by cosine of 2 X. Plus 2 multiplied by the sign of 2 X. In quantity. So taking this derivative, this is going to be equal to. When you take the derivative with respect to X over a first term of 2 multiplied by cosine of 2 X, we'll have 2, and here we're taking a derivative of a cosine function, so we'll get a minus sine function. So we'll have 2 multiplied by minus sine of 2 X. And we'll still have to apply our chain rules so we need to multiply this by the derivative with respect to X of 2 X, which is 2. Then we'll have plus, and when we take the derivative of the second term, we'll have the derivative with respect to X of 2 multiplied by sin of 2 X. So that will give us 2 multiplied by the derivative with respect to X of sin of 2 X. And so here we're taking a derivative of sine function, which will give us a cosine function. So we'll have cosine of 2 X. And again, we need to apply our chain rule and multiply it by this, multiply this by the derivative with respect to X of 2 X, which is 2. So simplifying this expression, this is going to be equal to 4, multiplied by cosine of 2 X. Minus or multiplied by sin of 2 X. And so now what we need to do is evaluate our second derivative here at each of our critical points. So we need to calculate F of 3 pi divided by 8. And this is going to be equal to or multiplied by cosine of the quantity of 2 multiplied by 3 pi divided by 8 in quantity, minus 4 multiplied by the sign of the quantity of 2 multiplied by 3 pi divided by 8 in quantity. And when we evaluate this expression, this is going to be equal to -4, multiplied by the square root of 2. So here we see our second derivative is negative at this critical point. And we call that if the second derivative is negative at this point, that means that we have a local maximum. We have a local maximum. At X equal to 3 divided by 8. So now we need to evaluate F at our other points. So we need to evaluate F of 7 pi divided by 8. And this is going to be equal to 4 multiplied by the cosine of the quantity of 2 multiplied by 7 pi divided by 8 in quantity, minus 4 multiplied by the sine. Of the quantity of 2 multiplied by 7 pi divided by 8. In quantity. And when we evaluate this expression, this is going to be equal to 4, multiplied by the square root of 2. So here we see our second derivative is positive at X equals to 7 pi divided by 8. So that means that we have a local minimum at this point. We have a local minimum. At X equal to 7 pi divided by 8. So, we do have local extrema at our two critical points. So now, what we need to do is evaluate our function F of X at each of these points. And so we're gonna look at F of 3 pi divided by 8. And this is going to be equal to. A sign Of the quantity of 2 multiplied by 3 pi divided by 8. In quantity minus cosine. Of quantity of 2 multiplied by 3 pi divided by 8 in quantity. And when we evaluate this expression, we get The square root of 2. So next we need to evaluate F of 7 pi divided by 8. And this is going to be equal to the sign. Of the quantity of 2 multiplied by 7 pi divided by 8 in quantity minus cosine of the quantity of 2 multiplied by 7 pi divided by 8. In quantity. And when we evaluate this expression, this is going to be equal to minus the square root of 2. So for our final answer, we're going to have a Local minimum. At X equal to 7 divided by 8, with F of 7 pi divided by 8. Be equal to minus square root of 2. And we'll have a local maximum. When X is equal to 3 pi divided by 8, with F of 3 pi divided by 8 equal to the square root of 2. So, those will be 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) = sin x − cos x,0 ≤ x ≤ 2π

Key Concepts

Local Extrema

Derivative

Interval Analysis

Watch next