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 2x, 0 ≤ x ≤ π

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to find the local extrema of the function GFT, which is equal to cosine of 3T. On the interval of 0 is less than or equal to T, which is less than or equal to pi divided by 2. So we need to find the local extrema of our function GFT. So for that, we'll need to find our critical points, and for that, we'll need to find our first derivative. So we'll need to calculate G of T, which is going to be equal to the derivative with respect to T. Of cosine of 3 T. So this is going to be equal to, and here we call that when you take the derivative of a cosine function, you get a minus sign function, so you'll have minus sign. Of 3T But we also need to apply our chain rules, so we need to multiply this by the derivative with respect to T of 3T, which is just 3. This means that D of T is equal to -3 multiplied by sine of 3 T. So, in order to find our critical points, those occur whenever our first derivative is undefined or equal to zero. Now, in our case here, The sine function is defined for all values of T, and so that means that G T is defined everywhere. So the only critical points are going to be due to our G T being equal to 0. So we're gonna set minus 3 multiplied by sine of 3 T. Equal to 0. Dividing both sides by minus 3, we'll get sine of 3 T. is equal to 0 and recall that the sine function is zero whenever its argument is an integer multiple of pi. So that means that 3 t is going to be equal to n multiplied by pi wheren here is an integer. So solving for T by dividing both sides by 3, we'll get T as equal to n multiplied by pi divided by 3. So now we need to find the values of T that fall within our interval. So, we'll start off by looking at N equal to 0, so we'll have T equal to 0 for that case. When we look at N equal to 1, that means that T is going to be equal to pi divided by 3. And if you look at the case where N is equal to 2, we'd have T is equal to 2 pi divided by 3. So that is greater than pi divided by 2. So, these two points are the only ones that fall within our interval. So, we need to do now is determine if each of these critical points is a local minimum or maximum. And to do this, we're we're going to be using the second derivative test. So we'll need to calculate G of T. And this is going to be equal to the derivative with respect to T of the quantity of minus 3 multiplied by sine of 3 T in quantity. So taking this derivative, this is going to be equal to -3, and that's going to get multiplied by the derivative with respect to T of sin of 3T. And recall that when you take the derivative of a sine function, you get a cosine function. So we'll have the cosine of 3T, but we'll need to apply a chain rule and multiply this by the derivative with respect to T of 3T, which is just 3. So simplifying this expression, this is going to be equal to minus 9, multiplied by cosine of 3 T. So now we need to do is calculate the value of our second derivative at each of our critical points. So, starting with T equals 20, we'll have G of 0. So this is going to be equal to -9, multiplied by cosine. Of 3 multiplied by 0. And when we evaluate that, that expression, that it's going to be equal to minus 9, since the cosine 0 is just 1. So here we see that G 0 is actually a negative value. So our second derivative is negative. So that means that we have a local maximum at T equal to 0. The T equals 0, we have a local maximum. Next, we'll calculate G of pi divided by 3. So this is going to be equal to -9, multiplied by the cosine. Of quantity of 3 multiplied by pi divided by 3 in quantity. And evaluate this expression, this is going to be equal to 9, since we'll have the cosine of pi, and the cosine of pi is -1, so -1 multiplied by -9 gives me 9. And here we see that our second derivative is positive. At this point, that means that we have a local minimum at this point. So we have a local minimum at T equal to pi divided by 3. So now we just need to calculate the value of our function. GFT at each of our critical points. So, G of 0 is going to be equal to. Cosine of the quantity of 3 multiplied by 0. Which is equal to one. And G of pi divided by 3 is going to be equal to cosine. Of the quantity of 3 multiplied by pi divided by 3. In quantity. And evaluate the expression that's gonna be equal to minus 1. So, for a final answer, we will have a Local minimum. At X At T equal to pi divided by 3, with G of pi divided by 3 equal to minus 1. And we'll have a local maximum. At T equal to 0, with G of 0. Equal to one. And 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) = sin 2x, 0 ≤ x ≤ π

Key Concepts

Local Extrema

Derivative and Critical Points

Trigonometric Functions and Their Derivatives

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