Use the second derivative test to find the local extrema of the given function.f(x)=4sinxcosxf\left(x\right)=4\sin x\cos xf(x)=4sinxcosx; 000 < xxx < π\piπ

Local max of 2 2 2 at x = π 4 x=\frac{\pi}{4} x = 4 π , local min of − 2 -2 − 2 at x = 3 π 4 x=\frac{3\pi}{4} x = 4 3 π

Use the second derivative test to find the local extrema of the given function. f ( x ) = 4 sin x cos x f\left(x\right)=4\sin x\cos x f ( x ) = 4 sin x cos x ; 0 0 0 < x x x < π \pi π

In this problem, we're asked to use the second derivative test to find the local extrema of the given function. Now the function we're working with here is f of x equals 4 times the sine of x times the cosine of x. And we're given a restriction on our domain where x has to be greater than 0 but less than pi. So we're working on an interval from 0 to pi where those endpoints are not actually included in the interval. Remember that to use our second derivative test, we first need to find where our first derivative is equal to 0. Once we have those values, we can plug them into our second derivative and then pay attention to the sign. The sign of that value plugged into that second derivative will tell us whether we're dealing with a local maximum or local minimum. Remember, when working with the second derivative test, we don't wanna be working with points where that first derivative does not exist or where that second derivative is equal to 0. If either of those things happen, we need to use the first derivative test. So let's go ahead and get started here by just finding that first derivative so we can find where it's equal to 0. So we're looking for f prime of x here because I have this 4 out front. I'm just gonna go ahead and pull that out and focus on the sine multiplied by the cosine. Because I have 2 functions being multiplied, I need to use the product rule to find this derivative. So remember our product rule memory tool, left d right plus a right d left. So I'm going to take that left hand function, sine of x, multiply it by the derivative of that right hand function, which is negative sine of x. Then I'm going to add that together with that right hand function, cosine of x, times the derivative of that left hand function, the derivative of sine is cosine. So if I go ahead and multiply this out, this gives me 4 times negative sine squared x plus cosine squared x. Now, here, I notice that I am dealing with a trig identity. Remember, when you're working with trig functions, you should always try to look and see if there's any identities that will help you simplify to make your life easier. Now here, I have a negative sine negative sine squared plus cosine squared. This is our cosine double angle identity. So I can rewrite this as 4 times the cosine of 2 x. This will make it easier to set this equal to 0. So setting this equal to 0 here, that 4 will cancel out. So I just wanna find where the cosine of 2 x is equal to 0. Now because this argument is not just x, that means that I need to take it and I need to set it equal to whatever makes the cosine 0. So what makes the cosine 0? Well, what makes the cosine 0 is pi over 2 plus n pi because we know that pi over 2, 3 pi over 2, and coming back around again and again always gives us a cosine that is 0. Now we wanna actually solve for x here by dividing by 2. Remember, dividing by 2 is the same thing as multiplying by 1 half. So when this cancels here, this gives us x is equal to pi over 4, multiplying that by 1 half, plus n pi over 2. Now remember, we are working within a restricted interval from 0 to pi. So we only want values of x that are within that interval. Pi over 4 is definitely within that interval. So I know that one of these values that makes the cosine 0 that is within this interval is pi over 4. Now if I add a one increment of pi over 2, that will give me 3 pi over 4, which is still within my established interval. So x can also be equal to 3 pi over 4. Now if I tried to add another factor of n pi over 2, that would mean that we are going outside of that interval. So we're only going to be working with these two values here because they are inside the interval from 0 to pi. So now that we actually have these values, we wanna take them and we wanna plug them into our second derivative. To do that, we just need to find our second derivative first. So f double prime of x. Now I can use the chain rule here because remember our first derivative is 4 times the cosine of 2x. So that's what we're differentiating here. So this is going to be 4 times the negative sign of 2x times 2, using the chain rule to get their derivative of that 2x. This gives us a negative 8 sine of 2x. Now we wanna take these values and plug them in and see whether they are positive or negative. So we wanna find f double prime of pi over 4 and f double prime of 3pi over 4. So we want to plug these values into our second derivative here. So that is negative 8 times the sine of 2 times pi over 4 and negative 8 times the sine of 2 times 3 pi over 4. And remember, we don't actually care what these values are, just whether they're positive or negative here. So we can think about whether these values are going to be positive or negative, or you can simply just multiply them out. Now if we actually simplify what's going on in the argument here, this becomes negative 8 times the sine of pi over 2, simplifying that. The sine of pi over 2 is just 1. So this is negative 8 times 1. That is a negative value. Because this is a negative value, that tells us that at x equals pi over 4, we are dealing with a local maximum. So we know that we have a local max here. And we can revisit that in just a second after we finish plugging in 3 pi over 4. Now this will end up working out to be negative 8 times the sine of 3 pi over 2. The sine of 3 pi over 2 is a negative one, so this ends up being a negative 8. Negative 8 times negative 1, this results in a positive value. Because this is a positive value, that tells us that this value, 3 at x equals pi over 4, gives us a local minimum. Remember, this is based on our second derivative test here. So we know where our local min and max are located here. Now we actually want to find what these values are by plugging these back into our original function f of x. So let's go ahead and do that here. In order to do that, we want to find f of pi over 4 and f of 3 pi over 4. Now remember, here's where we're plugging back into our original function. We're no longer working with our first or second derivative. So just a reminder of what our original function was, 4 times the sine of x times the cosine of x. So this is 4 times the sine of pi over 4 times the cosine of pi over 4. And then the same thing here, plugging in 3 pi over 4, let's just go ahead and write this out as well. Now this is a case where we actually do want the values. Typically, we're focused on the sine, but because we're in this last step, we actually want to find out what these minimum and maximum values are. 4 times the sine of pi over 4 times cosine of pi over 4. This gives us 4 times a root 2 over 2 times a root 2 over 2. Now this 4 will cancel both of those twos out. Then we just have a root 2 times root 2. This just gives us a positive 2. So based on this, this tells us that since this pi over 4 was a local maximum, that we have a local maximum of 2 atx equals pi over 4. So we have that local maximum value. Let's finish out here by finding our local minimum value. This is going to be 4 times root 2 over 2 times negative root 2 over 2. So we see that we're working with the same exact values here, except we have that negative. So this will work out to just be negative 2. So based on what we found above, we know that 3 pi over 4 is minimum. So based on this value, we now know that we have a local minimum value of negative 2 at x equals 3 pi over 4. So we have our final answer here that we have a local max value of 2 at x equals pi over 4 and a local minimum value of negative 2 at x equals 3 pi over 4, having used our second derivative test. Feel free to let me know if you have any questions here, and let's keep going.

Use the second derivative test to find the local extrema of the given function. f ( x ) = 4 sin ⁡ x cos ⁡ x f\left(x\right)=4\sin x\cos x f ( x ) = 4 sin x cos x ; 0 0 0 < x x x < π \pi π

Local max of 2 2 2 at x = π 4 x=\frac{\pi}{4} x = 4 π , local min of − 2 -2 − 2 at x = 3 π 4 x=\frac{3\pi}{4} x = 4 3 π

Local min of − 2 -2 − 2 at x = π 2 x=\frac{\pi}{2} x = 2 π , local max of 8 8 8 at x = π x=\pi x = π

Local max of 2 2 2 at x = π 2 x=\frac{\pi}{2} x = 2 π , local min of − 2 -2 − 2 at x = π x=\pi x = π

Local min of − 8 -8 − 8 at x = π 4 x=\frac{\pi}{4} x = 4 π , local max of 8 8 8 at x = 3 π 4 x=\frac{3\pi}{4} x = 4 3 π

5. Graphical Applications of Derivatives

The Second Derivative Test

Calculus

5. Graphical Applications of Derivatives

The Second Derivative Test

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Multiple Choice

Use the second derivative test to find the local extrema of the given function.
$f\left(x\right)=4\sin x\cos x$ ; $0$ < $x$ < $\pi$

Local min of $-2$ at $x=\frac{\pi}{2}$ , local max of $8$ at $x=\pi$

Local max of $2$ at $x=\frac{\pi}{2}$ , local min of $-2$ at $x=\pi$

Local min of $-8$ at $x=\frac{\pi}{4}$ , local max of $8$ at $x=\frac{3\pi}{4}$

Local max of $2$ at $x=\frac{\pi}{4}$ , local min of $-2$ at $x=\frac{3\pi}{4}$

Verified step by step guidance

First, recognize that the function given is f(x) = 4sin(x)cos(x). This can be simplified using the double angle identity: sin(2x) = 2sin(x)cos(x). Therefore, f(x) = 2sin(2x).

To find the critical points, calculate the first derivative f'(x). Using the chain rule, f'(x) = 4cos(2x). Set f'(x) = 0 to find the critical points, which gives cos(2x) = 0.

Solve cos(2x) = 0 for x within the interval 0 < x < π. The solutions are x = π/4 and x = 3π/4.

Next, calculate the second derivative f''(x) = -8sin(2x). Use the second derivative test to determine the nature of the critical points. Evaluate f''(x) at x = π/4 and x = 3π/4.

If f''(x) > 0 at a critical point, it indicates a local minimum. If f''(x) < 0, it indicates a local maximum. Use this information to identify the local extrema at x = π/4 and x = 3π/4.

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