# Homework Help: Tips on how to graph this one's derivative?

1. Feb 23, 2004

### ACLerok

one of my calc assignments asks me to graph the derivative of a graph which can be seen here: http://www.eden.rutgers.edu/~cjjacob/images/fifth.jpg [Broken]

i already know that when theres a cusp on the graph, the derivative of that point is at zero and that slope determines how the derivative acts. can anyone enlightnem me and give me tips on how to graph this one's derivative? Thanks.

Last edited by a moderator: May 1, 2017
2. Feb 23, 2004

### ShawnD

First of all, break the graph into sections where you see patterns. The graph starts off like a quadratic; make that a section. After that, the slope starts to decrease but stays positive; make that a section. After that, there is a vertical asymtote (slope is infinite) then the slope is negative but becomes less negative until the slope is 0; make that a section.

Just break it into pieces and it's quite easy.

Last edited: Feb 23, 2004
3. Feb 23, 2004

### HallsofIvy

"i already know that when theres a cusp on the graph, the derivative of that point is at zero"

Then you have a real problem. That's completely wrong. When there is a cusp on the graph, the derivative does not exist! It is when the graph is "level" (horizontal) that the derivative is 0.

ShawnD's advice is very good. From the left, the derivative starts out close to 0, increases and has an asymptote (goes to infinity) where the graph appears to be vertical. The slope is not negative after that (ShawnD was talking about the section after the first dotted line. I am still talking about the portion before it). It comes back down from "infinity" to some positive number at that dotted line. There is, of course, no derivative at the dotted line- the function shown is not even continuous there. On the right side of the dotted line, the derivative comes up from -infinity (the dotted line is a vertical asymptote), goes up to 0 where the curve "bottoms out" and the becomes positive. It has some positive value approaching that cusp, has no value at all (not zero!) at the cusp (show an "open" circle there) and the has some negative value immediately after. In fact, since the graph appears to be a straight line (with negative slope) the derivative graph should be a horizontal straight line (at some negative value). At the lower cusp, again an "open" circle showing no derivative and then a positive derivative. Almost horizontal- looks to me like the graph bends up just a little- suddenly increasing to a large positive value, then dropping back to 0 at the top of the "arch", then going to negative infinity as the graph approaches that second dotted line. An open circle at the dotted line (the function is not even defined there) and it looks to me like the function itself is a constant past that which means, of course, that the derivative is 0.

4. Feb 23, 2004

### ACLerok

ok i see.. so is there a general rule or method of graphing derivatives? like how does the slope of the graph or if its + or - relate to how the graph of the derivative will look like?