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.
5. y=x+sin(2x), -2π/3≤x≤2π/3

5. y=x+sin(2x), -2π/3≤x≤2π/3

Accepted Answer

Welcome back, everyone. In this problem, we want to determine the inflection points and intervals of concavity for the function Y equals 3X minus sine X where X is greater than or equal to negative pi, but less than R equal to pi. Now, if we're going to determine these inflection points and intervals of concavity, let's first ask ourselves, what do we know about inflection points? Well, recall that at the inflection point, the second derivative changes sign when it's equal to zero. So to find the inflection point, we need to find the second derivative of our function, set it to zero, and then figure out what's happening to the sign around those inflection points or around those values to tell if they are actual inflection points. OK? So let's try to figure that out. First, let's solve, OK, for the second derivative. And set it to 0. Now we know that our function is Y equals 3X minus sine X. So if we find our first derivative, OK, that means we're going to differentiate 3 X, which equals 3, and then we're going to differentiate sin X, which equals the cosine of X. So the first derivative is 3 minus the cosine of X. To find a second derivative, we differentiate that again. So now, we're gonna find the derivative of 3, which is 0, and the derivative of negative plus X, which is sin X, OK? So our second derivative is sin X. Know that we're going to set it to 0, OK. That means that at the potential inflection point. OK. At the potential inflection points. Our second derivative, the sine of X will be equal to 0. Now what value of X will make the sign of X equal to zero within our interval? We'll notice that our interval is from negative pi to pi. So this implies that X is going to be equal to either negative pi 0, or pi because the sign of negative pi is 0, the sign of 0 is 0, and the sign of pi is 0. So these are potential inflection points. No, we can check if the second derivative changes signed at these values to tell if they are actual inflection points because essentially what we're saying here, if we were to think about this on a number line, OK, so we have negative pi 0 pi, OK, and we have negative and positive infinity on both ends, then now we're saying what's happening to our sign before and after these points. Once it changes sign, then we know for sure that it's an inflection point. Now, for X, OK, let me put that in red. So now we're testing. Testing if they are, or testing the sign of the second derivative. At these points, OK, then. Notice that For X inside uh. Uh between the interval from negative pi to 0, OK. Then the sign of X. Is going to be negative, OK? So here we know the sin of X will be negative within this interval. On the other hand, for X in the interval from 0 to pi, OK, then we know that the sign of X is positive, OK? And so that tells us then that since it's negative in this interval and positive in this one, there is a change in sign at X equals 0, meaning X equals 0 is an inflection point. However, X equals negative pi and pi do not change sign, so they are not inflection points. So we can say that X equals 0 is an inflection point. So we figured out the first part of our problem. No, we need to figure out the second point that is to determine the intervals of concavity. How can we do that? Well, recall that if the second derivative is positive, the function is concave up, and if it is negative, the function is concave down. So now, Here notice that Or or second derivative. Which is sin X, OK. On The interval from negative pi to 0, OK. Like we said before, we know that the sine function is negative. So that means our function Y of X is concave down. OK? Because our second derivative is negative. On the other hand, On the interval from 0 to pi, OK, we know that the sine function is positive. So that means, or the second derivative is positive, so that means our function is concave up. Thus X equals 0 is our inflection point and our function is concave down on the interval from pi pi to 0, while it is concave up on the interval from 0 to pi. 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.
5. y=x+sin(2x), -2π/3≤x≤2π/3

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.5. y=x+sin(2x), -2π/3≤x≤2π/3

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.
5. y=x+sin(2x), -2π/3≤x≤2π/3