For this problem, we want to find the critical points of a few functions determine whether or not whether they are max men or neither and determine whether or not the tangent line will be parallel to the horizontal axis at that critical point. So our first function is negative six X cubed plus 18 X squared plus three. And that is labeled as for this problem one moment. That is labeled as H. Of X. So taking H prime of acts, we will get just 36 x minus 18 X squared having that. Uh we have a common factor of uh in fact it's a common factor of 18 X between the two terms. So we can actually factor out a negative 18 X and write this as negative 18 x times x minus two. Which tells us that we'll have a critical point at X equals zero and a critical point at X equals two. Now, since we have a negative leading coefficient on an odd polynomial, that means that will essentially have double negatives when x is less than zero. So we'll be coming in positive to the left of X equals zero. Then turning too negative in between zero and two. Then turning to positive again. Now that means that we'll have, oh excuse me. You should be careful here. We have on our derivative. We have a negative leading coefficient on an even uh even um polynomial, even degree polynomial. And so that actually means that we'll be coming in negative turning to positive then turning too negative again. So what that means going from negative to positive means that we have amen and going from positive to negative means that we have a max for the second part of the problem barbie we have GFT equals T to the power five first eq It means that G prime of T. Is going to equal five T to the power of four plus three. T squared can factor out the t squared from the two. We have t squared times five T squared plus three. This will only ever equal zero when T equals zero. So we have a critical point where T equals zero. Here, we have everything positive in an even degree polynomial, which tells us that we won't have a max or amen we'll be going from positive derivative to positive derivative, but we still have a critical point there at T equals zero. And because of the because of the fact that it is corresponding to a derivative equal zero critical point, that does mean that we will have a tangent line parallel to the horizontal axis there. For part C. We have the function why equals x minus five to the power of 1/5. Which then in turn means that y prime is going to equal by applying quotient rule will find that Y prime is going to be 1/5 times x minus five. The power of 4/5. Now this has no point where it will equal zero, but we do have a point where the derivative will be undefined when X equals five. So here it will, the tangent line will not be parallel to the horizontal axis, it will be perpendicular than for part D. We have the function F of X equals x squared minus one to the power of 1/3. That means then that f prime of X. Again, we're applying the in this case will be applying the chain world, we'll have to X over three times X squared minus one. The power of to over three. Now this will equal zero when X equals zero. So we have a critical point there. Oh, we'll also have that coming in, we'll have negative values. So we'll be switching from having negative derivatives to positive derivatives, which means that we'll have at X equals zero. That will be a minimum, and it will be parallel to the horizontal axis and we'll also have a critical point at X equals plus or minus one. Now those will be a critical points where we are perpendicular to the horizontal axis. The tangent line will just be a straight line there. So since they are perpendicular to the horizontal axis, those are neither maximum nor minimum.