Identify the inflection points and local maxima and minima of ...

Question

Identify the inflection points and local maxima and minima of the functions graphed in Exercises 1–8. Identify the open intervals on which the functions are differentiable and the graphs are concave up and concave down.
8. y = 2cosx - √2x, -π≤x≤3π/2

8. y = 2cosx - √2x, -π≤x≤3π/2

Accepted Answer

Welcome back, everyone. In this problem, we want to determine the inflection points and intervals of concavity for the function F of X equals 3 X plus 2 X quad, where X is greater than or equal to -2 pi but less than R equal to pi. Now let's focus on the first part of this problem. Let's try to find the inflection points for the function. What do we know? We recall that at the inflection points. OK. Then our second derivative, the second derivative of our function, in this case, the second derivative of F of X will be equal to 0. So what we'll need to do is to find the second derivative, set it to 0, and then try to figure out which values of X makes that possible. So let's go ahead and do that. Now we know that F of X equals 3 x plus 2 X2. So now that means then that the first derivative of F of X is going to be equal to the derivative of 3 X, which is 3 X, and the derivative of 2 X squared is equal to 4 X. So the first derivative is 3 X plus 4 X. Next, to find a second derivative, all we need to do is to differentiate the first derivative. I know the derivative of 3 + X is going to be equal to -3 sin x, while the derivative of 4 X equals 4. So the second derivative of our function is -3 X + 4. Know that we have the second derivative, we can set it to 0. So that means -3 X + 4 will be equal to 0, and if we subtract 4 from both sides and divide by -3, then we get the sin of X to be equal to 4/3. So now, we have to ask ourselves what value of X will make the sign of X be equal to 4/3. Well, let's think back to the domain of the sine of X, OK? Recall that the sin of X can only take values between -1 and 1. In other words, there's no value of X that will make the sign be greater than the absolute value of 1. So thus this tells us then that there is no solution to this equation, OK? And thus. There are no inflection points, OK? There are no inflection points. Where the second derivative of F of X will be equal to 0, OK? Next, let's move to the other part of our problem. Can we then figure out the intervals of concavity? Well, how do we do that? We can do that by analyzing the sign of the second derivative, OK? And remember we said that our domain is from negative to pi to pi inclusive. So within this interval, we need to determine the behavior of our second derivative. And no, because it's a continuous function, the concavity depends on the sign of -3 X + 4. So let's check the sign of the second derivative. At specific points on the interval interval, OK, so analyzing our second derivative. For a concavity. OK. And the interval from negative to pi to positive pi, OK. Then let's take the first point I'd say X equals 0. At X equals 0, then our second derivative is gonna be equal to -30 + 4. The sign of 0 is 0, so our result is going to be 4, which is positive. Now, let's take some other point. In our interval, OK, because now, since it's positive, this tells us that our function is going to be concave up. Let's say that we went to X equals negative pi. Well, now, our second derivative. OK, it's gonna be equal to F of negative pi, which equals -3 sin negative pi plus 4. The sign of negative pi is 0. So again, our answer is going to be 4, it's going to be positive, which means then that our function is also going to be concave up at X equals pi. Let's pick another value within our interval. Say, let's pick where X equals negative. Well, that was where X E equals negative pi. Sorry. So now let's actually use where X equals positive pi. In this case, then our second derivative, OK, is gonna be equal to -3 spi plus 4, which again is going to be equal to 4, which means that our function will be concave up. So what do you notice is happening here? Well, by now, you probably realize that the second derivative is always positive on the entire interval. So that implies the function is concave up on the entire interval. So, in summary, there are no inflection points because our second derivative never equals 0 within the given domain and the function is concave up on the entire interval. Thanks a lot for watching everyone. I hope this video helped.

Identify the inflection points and local maxima and minima of the functions graphed in Exercises 1–8. Identify the open intervals on which the functions are differentiable and the graphs are concave up and concave down.
8. y = 2cosx - √2x, -π≤x≤3π/2

Key Concepts

Inflection Points

Local Maxima and Minima

Concavity and Differentiability

Watch next

Identify the inflection points and local maxima and minima of the functions graphed in Exercises 1–8. Identify the open intervals on which the functions are differentiable and the graphs are concave up and concave down.8. y = 2cosx - √2x, -π≤x≤3π/2

Key Concepts

Inflection Points

Local Maxima and Minima

Concavity and Differentiability

Watch next

Identify the inflection points and local maxima and minima of the functions graphed in Exercises 1–8. Identify the open intervals on which the functions are differentiable and the graphs are concave up and concave down.
8. y = 2cosx - √2x, -π≤x≤3π/2