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.
4. y=9/14x^(1/3)(x^2-7)

4. y=9/14x^(1/3)(x^2-7)

Accepted Answer

Welcome back, everyone. In this problem, we want to find the inflection points of the function Y equals 4/3 of X3 multiplied by X2 minus 9. We also want to determine its intervals of concavity. Let's focus on the first part of our problem. Let's try to find the inflection points. What do we know about the inflection points of our function? Will recall that at the inflection points. OK. Then the second derivative of our function is either going to be equal to 0 or it's going to be undefined. So for us to figure out the inflection points, let's differentiate our function twice, set it to 0, and then solve for the values of X that make it 0, or for the values of X that will make it undefined. Now we can simplify our function a little bit further to help us differentiate. Now if we expand our term here, Y is going to be equal to 4/3 of x3 multiplied by X2, OK, or 4/3 of X3 + 2 we're multiplying x terms, so we add their powers. Minus 4/3 of X3 multiplied by 9. I know when we go ahead and simplify here. Then we should get this to be equal to 4/3 of X 7/3 minus 1212 multiplied by X to the 4 of 3. No, it makes it a bit easier to find our first derivative because this tells us then that the first derivative is going to be equal to 4/3 multiplied by 7/3 multiplied by X of 7/3 minus 1, which is 4/3. Minus 12 multiplied by 3. Multiplied by x of 1/3 minus 1 is negative 2/3. OK, so we've differentiated both terms by using the poor rule of differentiation. And now when we simplify, we'll get that to be 289 of X to the power of 4/3 minus 4 x 2/3. Now that we have the first derivative, we can differentiate it again to get the second derivative. So now this means that our second derivative is gonna be equal to 289. Multiplied by 4/3. Of X to the poor of 4/3 minus 1 1/3. Minus 4 multiplied by 2/3, OK, multiplied by X to the power of -5/3. I know when we go ahead and simplify here, OK, then we're going to get this to be equal to 112 divided by 27 X3. Plus 8/3 of X to the power of -5/3. This is the 2nd derivative of our function Y. Now remember what we had said earlier? At the second derivative, or at the inflection point, the second derivative will either be equal to zero or undefined. So let's try to figure out the value of X in both of those cases. At the point where the second derivative is equal to 0. Then we can set our expression or set our equation to 0. So all of this will be equal to 0. If we multiply through by 27, so let's put this here, multiply our entire equation. As a matter of fact, let me put that in red. So if we multiply through our entire equation by 27 X to the power of 5/3, it's going to eliminate our denominators and our negative exponents, OK? So now, when we do that, Then we should get 112 x squared plus 72. To be equal to 0 and if that's the case, then 112 x 2 is gonna be equal to negative 72 but we've run into a problem here. We have an Xquared term and Xsquared is always non-negative, so that means there are no real solutions to the second derivative of Y being equal to zero because we know that if X quad is here, then its result must always be a positive term. So now in this case. We can say that here there are no real solutions. Now how about our second scenario where our second derivative is undefined. Well, let's ask ourselves, what would make our second derivative be undefined? What would make this term, this entire function or this expression be undefined? Well, it would be undefined. Where x to the power of -5/3. Would be equal to 0, which would imply that X would be equal to 0, OK? Because here that means in our denominator, we'd be dividing by 0 and when we divide by 0, our result is undefined. However, we call that the original function Y is defined as X equals 0, so we'll have to check the behavior around X equals 0. So let's analyze its concavity, OK? So let's look at the second derivative just to the left and the right of X E equals 0. Now if we think about it here, when x approaches 0 from the right, OK, then X to the power of a third will approach 0 and X to the power of -5/3 will approach positive infinity. So that means our second derivative is going to approach positive infinity. And since its sign is positive, that means it will be concave up approaching 0 from the right. However, When it's approaching 0 from the left, then X to the of a third will approach 0, while x to the pore of negative/3s will approach negative infinity. So that means our second derivative will be approaching negative infinity as X approaches 0 from the left, which means that it's concave down. No, what that means is that we can tell the concavity changes from concave down to concave up as X passes through 0, and thus, that means X equals 0 is an inflection point, OK? So we've been able to find it by analyzing the sign of our second derivative around X equals 0. Now that we know it's an inflection point, we can determine the intervals of concavity, and to do that, we can analyze the sign of the second derivative where x is not equal to 0. So now, let me put this in red here. Analyzing concavity. Analyzing concavity, OK. By looking at the sign of the second derivative, it might make it easier to look at it if we have it written as one term. So now we know that we have the 2nd derivative as 112 divided by 27 X to the 3rd. Plus 883. Of X to the power of negative 5/3, OK. So now, if we combine both of these, we can write this as 112 X2 + 72 divided by 27 X of 5/3. And if we factored it, that's gonna be 8 multiplied by 14 X2 + 9 divided by 27 X to the power of 5/3. Now what can we tell about our sign here? Well, notice that our numerator has a squared term, OK? So now, that tells us that our numerator, 14 X2 + 9 is always going to be positive for all X, OK? So since this expression will be positive or a numerator will be positive. I know, since that's the case then for our denominator. Next to the pore of 5/3, it has the same sign. Has the same sign as X, whatever X is X to the 5/3 will have the same sign. So thus that tells us then for positive values of X or second derivative will be positive because a numerator will be positive and X will be positive. So thus that tells us our function will be concave up. And for X for negative values of X, our second derivative would also be negative because our numerator will be positive or denominator will be negative and positive divided by negative is negative. So that means our function will be concave down for values of X less than 0. So now, for a final answer, this tells us then. That therefore our inflection point. My inflection point is X equals 0. And for our intervals of concavity, our function is so and so let's say here, Y of X is concave down. And the interval from negative infinity to zero or closed parentheses and concave up. On the interval from 0 to infinity open closed parentheses. 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.
4. y=9/14x^(1/3)(x^2-7)

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.4. y=9/14x^(1/3)(x^2-7)

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.
4. y=9/14x^(1/3)(x^2-7)