Identify the local minimum and maximum values of the given function, if any.f(θ)=sinθ+cos2θf\left(\theta\right)=\sin\theta+\cos^2\thetaf(θ)=sinθ+cos2θ on [0,π] [0,π][0,π]

Local maxima of 5 4 \frac54 4 5 at x = π 6 & 5 π 6 x=\frac{\pi}{6}\&\frac{5\pi}{6} x = 6 π & 6 5 π , Local minimum of 1 1 1 at x = π 2 x=\frac{\pi}{2} x = 2 π

Identify the local minimum and maximum values of the given function, if any. f ( θ ) = sin θ + cos 2 θ f\left(\theta\right)=\sin\theta+\cos^2\theta f ( θ ) = sin θ + cos 2 θ on [ 0 , π ] [0,π] [ 0 , π ]

In this problem, we're asked to identify the local minimum and maximum values of the given function, if there are any. Now the function we're given here is f of theta equals sine theta plus cosine squared of theta, and this is specifically on the interval from 0 to pi. Now remember that to find our local minimum and maximum values, we can use the first derivative test, which tells us that once we find our critical points, we can test the sign of our derivative on either side of those critical points and observe the sign changes. This tells us whether we're dealing with a local maximum or a local minimum or neither of these. So let's go ahead and get started here by first finding those critical points, which means that we first need to find our derivative so that we can find where it's equal to 0 or does that exist. So finding our derivative here, that's f prime of theta. Now the derivative of sine is just cosine, so this is cosine of theta. And then we're gonna use the chain rule here. We can think of cosine squared of theta as cosine theta squared if that helps you to apply the chain rule here. So this is gonna be 2 times the cosine of theta, and then we wanna take the derivative of cosine, which is negative sine. So this is going to multiply by negative sine. Now we can do some simplification here. Because both of these have a cosine in common, I can factor that out. That gives me cosine of theta times 1 minus 2 sine of theta because this is a negative sign here. So this is our derivative here. Now within our inter within our interval from 0 to pi, there's nowhere this derivative won't exist, but we can go ahead and set it to 0 to get our critical points. So in doing that, since I have these two factors, that means I'm effectively setting each of these individual factors equal to 0. So I have cosine of theta equals 0 and 1 minus 2 sine of theta equals 0. So looking at this cosine theta equals 0, we wanna think, okay. What value of theta makes the cosine 0? Remember that we are specifically working on this interval from 0 to pi, so there is actually one value of theta for which this happens within that interval, and that is pi over 2. So we have our first critical point here, theta equals pi over 2. Then 1 minus 2 sine of theta, let's go ahead and move that that 2 sine theta to the other side, and then we'll swap them just to make both sides positive. That's 2 sine theta equals 1. Now if I divide both sides by 2, that is sine of theta is equal to 1 half. So we wanna find where for what value of theta is the sine equal to positive one half. Now thinking about this specifically for that interval again from 0 to pi, there are 2 values for which that happens, and that is theta equals pi over 6, and theta equals 5pi over 6. That is still within my interval from 0 to pi. So I have these 2 critical points here as well as this other one here. So with these 3 critical points in mind, I can go ahead and make my sign chart intervals here, keeping in mind that I am working specifically within the total interval from 0 to pi. So marking these critical points, I can go ahead and mark that pi over 6. That's my smallest critical point. Then pi over 2 and 5 pi over 6. Now when we create these intervals, we do wanna keep in mind that we have endpoints here. So this first interval goes from 0 to pi over 6 with that 0 included because it is an endpoint. We know it's included based on the information given to us in the problem. Then we have pi over 6 to pi over 2. This is, of course, an open interval with those critical points. Then from pi over 2 to 5 pi over 6, then finally from 5pi over 6 to that other endpoint, which is pi. Pi is included in that interval because it is an endpoint. We are given that information in the problem. Okay. We have our sign chart intervals. Now we wanna go ahead and choose a value within each interval to plug into that derivative that we found. Remember, we wanna choose these values such that they're easy to work with. Here, since we are dealing with trick functions, we typically wanna choose values that are on the unit circle, values that we know by memory. So here from 0 to pi over 6, since 0 is included in that interval, I'm just gonna go ahead and choose 0 itself. Then between pi over 6 and pi over 2, we can go ahead and choose pi over 3. Then from pi over 2 to 5 pi over 6, I'm gonna go ahead and just choose 2 pi over 3 for another familiar value. And then, again, here, since pi is included in that interval, I'm gonna go ahead and choose pi for that test value of that last interval. Okay. So we have a lot to plug in here. Let's go ahead and get going on this algebra. We're plugging these back into our derivative. That means that we want to find here f prime of 0, f prime of pi over 3, f prime of 2 pi over 3, and finally, f prime of a pi. So we're plugging these into our derivative. Remember that our derivative is right here. It is the cosine of theta times 1 minus 2 times the sine of theta. So that's what we're plugging this into. So setting that up for 0 here, that's the cosine of 0 times 1 minus 2 times the sine of 0. So when we calculate this out, we know that the cosine of 0 is going to be 1. And then 1 minus 2 times the sine of 0, that's actually also gonna give me 1 because this is 0. So this just gives me 1 times 1, which is one. We know this is a positive value. So on this interval, we know that our derivative is positive. Okay. For our next value, pi over 3, we wanna take the cosine of pi over 3, multiply that times 1 minus 2 times the sine of pi over 3. Now here we can think about whether these values are positive or negative based on their location on the unit circle. That can be helpful to us here. So I can go ahead and think about it that way. I know that the cosine of pi over 3 is going to be a positive value. I can go ahead and actually work that out. From my knowledge of the units are going to know that is 1 half. Then I have 1 minus 2 times the sine of pi over 3. This is going to work out to be 1 minus the square root of 3 because that 2 will end up canceling with that 2 on the bottom there. Then from here, we can see that since root 3 is a greater number than 1, we don't have to worry about what this value actually is. Remember, we really just care about its sign. We know that this will work out to be a negative value based on what we're working with here. So on this next interval, that derivative is negative. Now we can take the same approach for this next one. This is cosine of 2 pi over 3 times 1 minus 2 times the sine of pi over 3. Now when we do this here, this ends up being negative one half, but again times 1 minus root 3. So this is a negative times a negative. This results in a positive. Then finally, for Pi plugging in Pi, cosine of Pi times 1 minus 2 times the sine of Pi. Cosine of Pi is a negative one. 1 minus 2 times the sine of pi will give me a positive one because this is 0. This ends up being a negative. We know that it's negative one since we're just working with ones there. So that 3rd interval derivative is positive. 4th interval derivative is negative. So we can see that all of these different values do result in sign changes. Now here's where we wanna apply our first derivative test in step 3 and look exactly at what sign changes are happening here. Now we can see that at this critical point, pi over 6, that derivative goes from being positive, our function increasing, to then being negative, our function decreasing. So we know that this indicates a local max at this point, pi over 6. Then at pi over 2, we can see that our derivative goes from being negative, our function decreasing, to then being positive, our function increasing. And this indicates a local minimum at this value. And then finally at pi at 5pi over 6 at x equals 5pi over 6, our function goes from positive derivative, so increasing, to a negative derivative. Now it's decreasing. So this indicates a local maximum. Now since here we do wanna find what these values are, we are not quite done yet here. We need to go ahead and plug these back into our original function. So we want to find f of pi over 6, f of pi over 2, and f of 5 pi over 6 just to fully get our answer here. So let's go ahead and work that out. So coming all the way down to the bottom here to plug these values in, we are finding f of pi over 6. We know that this is a local maximum. F of pi over 2, we know this is a local minimum. And f of 5 pi over 6, this is another local maximum. Now remember, at this point in step 4, we are plugging this back into our original function, not our derivative. I know we've been working with the derivative the whole time. This is where we wanna go back to that original function, which here is the sine of theta plus the cosine squared of theta and plug those values into that. So here we wanna find the sine of pi over 6 plus the cosine squared of pi over 6. I'm going to go ahead and write this out for all of these. Sine of pi over 2 plus the cosine squared of pi over 2, then the sine of 5 pi over 6 plus the cosine squared of pi, 5 pi over 6. Now focusing in on that first value here, if we go ahead and just plug this into our calculators, you can also just work this out by hand since these are familiar values from the unit circle. And either way that you go about this, you should end up getting a value of 5 over 4 here. Then when we plug pi over 2, and again, you can work this out by hand since this is a familiar trig value, you should end up getting here just 1 because this is 1+0. Then for that 5 pi over 6, we actually end up with the same exact value that we got for our other local max, which we know happens commonly when dealing with trig functions. This is again 5 4ths. So if we wanna go ahead and state these full answers, these two values are local maxes. So we have a local maximum values of 5 4ths at x equals pi over 6 and x equals 5 pi over 6. Then we have a local minimum value of 1 at x equals pi over 2. Remember that we would actually have more local maxes and mins for these trig functions, but we were given an interval from 0 to pi with those endpoints. So I hope this helped you out, and I'll see you in the next one.

Identify the local minimum and maximum values of the given function, if any. f ( θ ) = sin ⁡ θ + cos ⁡ 2 θ f\left(\theta\right)=\sin\theta+\cos^2\theta f ( θ ) = sin θ + cos 2 θ on [ 0 , π ] [0,π] [ 0 , π ]

Local maxima of 5 4 \frac54 4 5 at x = π 6 & 5 π 6 x=\frac{\pi}{6}\&\frac{5\pi}{6} x = 6 π & 6 5 π , Local minimum of 1 1 1 at x = π 2 x=\frac{\pi}{2} x = 2 π

Local maxima of 5 4 \frac54 4 5 at x = π 6 x=\frac{\pi}{6} x = 6 π , Local minimum of 1 1 1 at x = π 2 x=\frac{\pi}{2} x = 2 π

Local maxima of 5 4 \frac54 4 5 at x = π 6 x=\frac{\pi}{6} x = 6 π , Local minimum of 1 1 1 at x = π 2 & 3 π 2 x=\frac{\pi}{2}\&\frac{3\pi}{2} x = 2 π & 2 3 π

5. Graphical Applications of Derivatives

The First Derivative Test

Calculus

5. Graphical Applications of Derivatives

The First Derivative Test

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Multiple Choice

Identify the local minimum and maximum values of the given function, if any.
$f\left(\theta\right)=\sin\theta+\cos^2\theta$ on $[0,π]$

Local maxima of $\frac54$ at $x=\frac{\pi}{6}$ , Local minimum of $1$ at $x=\frac{\pi}{2}$

Local maxima of $\frac54$ at $x=\frac{\pi}{6}\&\frac{5\pi}{6}$ , Local minimum of $1$ at $x=\frac{\pi}{2}$

Local maxima of $\frac54$ at $x=\frac{\pi}{6}$ , Local minimum of $1$ at $x=\frac{\pi}{2}\&\frac{3\pi}{2}$

No local extrema

Verified step by step guidance

First, find the derivative of the function f(θ) = sin(θ) + cos^2(θ) with respect to θ. Use the chain rule for the term cos^2(θ).

Set the derivative equal to zero to find the critical points. Solve the equation to find the values of θ within the interval [0, π].

Evaluate the second derivative at each critical point to determine the concavity. This will help identify whether each critical point is a local minimum or maximum.

Check the endpoints of the interval [0, π] by evaluating the original function f(θ) at θ = 0 and θ = π to ensure no local extrema are missed.

Compare the function values at the critical points and endpoints to determine the local maxima and minima within the interval [0, π].

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