The limit is H approaches zero of the quantity X + H ^3 / 4 -, X ^3 / 4 all over H So the limit is H approaches zero. We have to foil out that X + H ^3, multiply it, and then multiply it again. So we get X ^3 + 3 X squared H + 3 XH squared plus H ^3 -, X ^2 -, X ^3 all over 4H so that the X cubes then cancel and we get three X ^2 + 3 XH squared plus H ^3 / 4 H. And the HS will cancel in each and every term. So we get three X ^2 + 3 XH plus H ^2 / 4. So the derivative when we stick in at H going to 0 is just going to be three X ^2 / 4. Now the second part says to graph the original function, which is a cubic function X ^3 / 4 with the positive coefficient. So no, we know the end behavior has got to have as X goes to Infinity, Y is going off to Infinity, and as X is going to negative Infinity, Y is got to go to negative Infinity. Now this is the derivative and the derivative is a quadratic. So it's a parabola with a positive coefficient. So the parabola has got to be going up the X as it goes to Infinity, the Y has got to go to Infinity, and X as it goes to negative Infinity, Y has got to go to Infinity. So part C says determine the value when the first derivative is positive, while the first derivative is always positive except at the origin where it's equal to 0 and it's not negative anywhere. So that would be our choice. A, our question D says when is the original function increasing and decreasing, Increasing going up. So if we look at this cubic equation, we're going up from negative Infinity to zero. At 0, we're not doing anything one way or another. And at 0 to Infinity, we're going up again. So the big thing on this problem is that we're supposed to understand that if it's a positive slope that our derivative is positive. IE if the original function is increasing and has a positive slope, then the derivative is going to be positive. Or the Y values here are positive. Right here. The derivative or the original function wasn't increasing or decreasing, hence it had a 0 derivative there.