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) = csc²x − 2cot x, 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 H of X, which is equal to sin squared Xs, plus 2 multiplied by cosine of Xs on the interval, 0 is less than X, which is less than 2 pi. So we need to find the local extrema of our function H of X. And so to do that, we need to find the critical points, and for that, we will need our first derivative. So we'll need to calculate H of X, and that's going to be equal to the derivative with respect to X of quantity of sin squared of X. Plus 2 multiplied by cosine of X in quantity. So, taking this derivative, this is going to be equal to. When I take the derivative of the first term, which is sin squared of X with respect to X. We'll need to use both our power rule as well as our chain rule. So applying the power rule first, we'll have 2 multiplied by sine of X. Then we'll need to apply our chain rules, so we need to multiply this by the derivative with respect to X of sin of X, which is cosine of X. Then we'll have plus 2 multiplied by the derivative with respect to x of cosine of X, and recall that that is going to be equal to minus sine of X. We can simplify this expression by factoring out a 2 multiplied by sin of X out of each of the terms. This is going to be equal to 2 multiplied by sine of X, multiplied by the quantity of cosine of X minus 1 in quantity. And so now we need to determine our critical points and recall that critical points are the values of X at which our first derivative is either undefined or equal to 0. And since our first derivative is made up of sine and cosine functions, those are going to be continuous everywhere and therefore our function H X is going to be defined everywhere. So the only time that we can have critical points is when our first derivative is equal to 0. So we're going to set. 2 multiplied by sine of X, multiplied by the quantity of cosine of X minus 1 in quantity equal to 0. So we're going to set each one of these factors equal to 0 and solve for x. So we'll have for the first factor we'll have 2 multiplied by sin of x equal to 0, and therefore sine of X is equal to 0, and recall that sin of x is equal to 0 at integer multiples of pi. So at 0, i, 2 pi, and so forth. However, the only value of X that is contained in our interval, which is the open interval from 0 to 2 pi, so we're not including in points, is going to be the value when X is equal to pi. So that's gonna be the solution for the first factor. Now for the second factor we're going to have. Cosine of X -1 equal to 0. And solving this for X, we'll have to add 1 to both sides. We'll have cosine of X equal to 1, and we call that cosine of X is equal to 1 at integer multiples of 2 pi. So at 024, and so forth. However, none of those values are within the open interval from 0 to 2 pi. And so we don't have any real solutions for this second factor on the open interval from 0 to 2 pi. So the only critical point that we need to look at is when X is equal to pi, and we need to determine if we have a local maximum or a local minimum at that point. So to do that, we're going to use the second derivative test. So we're going to calculate H double prime of X in order to do that, and this is going to be equal to the derivative with respect to X. Of the quantity of 2 multiplied by sine of X, multiplied by the quantity of cosine of X minus 1 in quantity. So when we take this derivative, we're going to need to apply our product rule. So we call for the product rule that we're going to have the derivative of the first term. So we'll have the derivative with respect to X of 2 multiplied by sin of X, and so that'll give me 2 multiplied by cosine of X. Then that gets multiplied by the second terminal product, which is the quantity of cosine of X minus 1 in quantity. Then we'll have plus the first term in a product, which is 2 multiplied by sin of x. Multiplied by the derivative of the second term in a product. So we'll have the derivative with respect to X of the quantity of cosine of X minus 1, and we take that derivative, we'll get minus sine of X. And so we can simplify this expression, so we'll factor out of 2 out of each of the term, and then multiply together all the other terms. So this is going to be equal to. 2 multiplied by the quantity of cosine squared of X minus cosine of X. Minus sine squared of X. In quantity. And so now we need to um Evaluate this second derivative when X is equal to pi. We'll have a prime of pi. That's gonna be equal to 2 multiplied by the quantity of cosine squared of pi minus cosine of pi. Minus sine squad of pi in quantity. And so evaluating this expression, this is going to be equal to 4. And so here, our second derivative is positive when X is equal to pi. And so since our second derivative is positive, that means that we have a local minimum at this point. We will have a local minimum. Add x equal to pi. And we don't actually have any local maxima, so since we only had one critical point to evaluate. So, we need to determine the value of H of X at our critical point. So, we'll have H of pi. And that's going to be equal to sin squared of pi. Plus 2 multiplied by cosine of pi, and when we evaluate this expression, this is going to be equal to -2. So for a final answer, we're gonna have a local minimum. At X equal to pi. With each of pi. Equal to -2. And for local. Maxima We actually have none. So those are the answers that we were looking for for this problem. 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) = csc²x − 2cot x, 0 < x < π

Key Concepts

Trigonometric Functions

Derivatives and Critical Points

Interval Analysis

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