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) = x / 2 − 2sin (x/2), 0 ≤ x ≤ 2π

Accepted Answer

Welcome back, everyone. In this problem, we'd like to find the local extrema of the function G of T equals root 3 T minus 4 cost of half of T on the interval where T is greater than or equal to 0 but less than or equal to 4 pi. Now to find the local Xtrema, we need to find the critical points, determine whether they are maximum or minimum points, and then find that maximum or minimum value. So let's first start by finding the critical points of our function. We know that at the critical points, the first derivative of our function G of T. Is going to be equal to 0, so we need to find the derivative of g of T. Now to do that, we need to differentiate route 3 T and -4 cast half of T. And when we do that, the derivative of root 3T is going to be equal to root 3, while the derivative of -4 cast half of T equals + 22 of half of T. OK? So this is our first derivative. And we said that at the critical points this value is going to be equal to 0. Now, if the first derivative is equal to 0, OK, then this would imply if we subtract root 3 from both sides and divide by 2, that the sign of a half of T equals negative root 3 divided by 2. No, recall, OK. That generally. If the sign of an anger theta is equal to negative root 3 divided by 2, then that means our angle theta is going to either be equal to 4/3 of pi plus 2n pi or it's going to be equal to 5/3 of pi plus 2n pi for N as an integer. In other words, if the sign of our angle equals negative root 3 divided by 2, then our angle is going to be 4/3 of pi or 5/3 of pi plus a multiple of 2 pi. So with that idea. This would imply then. That's right here thus. This means that our angle half of T is going to be equal to those values 4/3 of pi plus 2n pi or it's going to be equal to 5/3 of pi plus 2 and pi. No, if we were to solve for T, this means then if we multiply through by 2, T is going to be equal to 8/3 of pi plus 4 N pi, OK? Or It's going to be equal to 10/3 of pi plus 4 in pi. Now what are our values going to be? Well, recall that we know that the value of T is going to be in the interval from 0 to 4 pi inclusive. So let's try to figure out the values of T using some of those multiples. First, let's start when N equals 0. Now when N equals 0, that means T is going to be equal to 8/3 of pi, or it's going to be equal to 10/3 of pi. Something interesting happens as we increase the number of multiples. For example, let's say that N E equals 1. Then no, T would be 1/3 of pi plus 4 pi or 10/3 of pi plus 4 pi, but that's going to carry it out of the interval where 4 pi would be the highest value. Thus, there are two values of T for critical points. Either T equals 8/3 of pi or T equals 10/3 of pi. Now that we've found the critical points, next we need to figure out their nature, and to figure out their nature, we can use the second derivative test because basically that test tells us that if we find the second derivative of our function and substitute the T values in this case for our critical points, if they are negative, if the second derivative is negative, it's a maximum, and if it's positive, then it's going to be a minimum. So let's first find the 2nd derivative and then apply the 2nd derivative test. Now remember that our first derivative was root 3 + 2 sine of half of T. So that means the second derivative G T is going to be equal to. The derivative of root 3, which is 0, plus the derivative of 2 multiplied by the sine of a half of T which is going to be equal to 2 or rather. Let's say a 1/2 multiplied by 2 costs 1/2 of tea. And that would simplify to the cosine of a half of T. So this is the second derivative of our function. Now that we have the 2nd derivative, we can apply the 2nd derivative test. Now that means we're gonna substitute our critical values into the second derivative. OK, derivative test. So let's do that. And now, let's start with when T equals 8/3 of pi. No, at T equals 83 of pie. Then this means that our 2nd derivative, OK. Is going to be equal to the cosine of 8/3 of pi multiplied by 1/2, which is going to be equal to 22 is less than 0. So since our second derivative is negative, this indicates that. When the point where T equals 8/3 of pi, OK, is a local maximum. And now that we know it's a local maximum, we can solve or figure out what that maximum value is by plugging this value of T into the original function. In other words, we can use, we can find G of 8/3 of pi. And no, that means remember our function G of T was route 3T minus 4 plus half of T. So this is gonna be route 3 multiplied by 8/3 of pi. Minus 4 multiplied by the cosine of 8/3 of pi multiplied by 1/2. I know when we work that out, you should find that it equals 8 root 3 pi plus 6 divided by 3. So T equals 8/3 of pies are local maximum and the maximum value is 8, root 3 pi plus 6 all divided by 3. Now let's test the next value of T, the next critical point. And that is when T equals 10/3 of pie. And when tea is 10/3 of pie, then the second derivative. It's gonna be equal to the cosine of 10/3 of pi multiplied by 1/2. I know when we go ahead and calculate here, OK. Then you should find that we get our 2nd derivative to be equal to 1/2. And a half is greater than 0. Since it's greater than 0, this tells us that our point T where T equals 10/3 of pi is a local minimum based on the second derivative test. I know, since it's a local minimum, we can find or try to figure out what that value is. And when we do that. Then this would imply that G of 10/3 of pi. Is going to be equal to Route 3 multiplied by 10/3 of pi. Minus 4 multiplied by the cosine. Of 10/3 of pi multiplied by 1/2, OK? And that is going to be equal to 10 root minus 10 root 3 pi. Minus 6 all divided by 3. So now we know that T equals 10/3 of pi is a low or in other words, there's a low minimum at T equals 10/3 of pi with a value of 10 root 3 pi minus 6 all divided by 3. Thanks a lot for watching everyone. I hope this video helped.

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) = x / 2 − 2sin (x/2), 0 ≤ x ≤ 2π

Key Concepts

Local Extrema

Derivative and Critical Points

Interval Analysis

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