Ari Stern:
  • A derivative does not exist at a cusp, hole, asymptote, vertical line, or a jump discontinuity


Sophia Hirsch: To make a more general statement, when the function is not continuous, the function is not differentiable.

Geneveive Tabby
A derivative also does not exist at a cornerpoint, if that wasn't said already. But to prove why this is true algebraically, we can use the formula of a derivative, which is:
external image moz-screenshot.jpgexternal image moz-screenshot-1.jpgformula_of_a_derivative.gif
(example as to why cornerpoints are not differentiable)
If, for example, we said that f(x)= IxI and looked at its derivative at x=0,
we would get two answers:

f(0+h) -f(0) / h = IhI- 0 / h = IhI / h
lim as h --> 0 from the right at IhI / h = 1 and lim from the left as h --> 0 IhI/h = -1

so, because the lim from the right and from the left don't match up at x=0, the derivative does not exist.
this is a picture of what a cornerpoint looks like:
picture_of_a_cornerpoint.gif

What are Cusps and why are they non-differentiable?
- To start off, what is a cusp? As seen in the photo below, a cusp is a type of function that is symmetrical over a line that passes through its "central axis". Also, The cusp's point can point either up or down.


external image moz-screenshot.jpgexternal image 600px-B%C3%A9zier_curve_with_cusp.svg.png

- Now, why are cusps non-differentiable ?
From the definition of a derivative, we know that the left and right hand limits approaching the x value in question must be equal. Now by looking at the graph above as an example, we can better see how we can use this part of the definition of a derivative to understand why cusps are non-differentiable. If one were to look at the values of the first derivative of our function f(x) as we moved over the interval (0,.45) one would see that the value of our first derivative is sharply decreasing as it becomes more and more negative. Therefore we can say that the limit of f(x) as x approaches .45 from the left is negative infinity. Now lets look at the other side. Here, if one looks at the value of the the first derivative approaching .45 from the right, we see that this value approaches positive infinity. This is why cusp's are non-differentiable. They break the rule that left and right hand limits must be equal when approaching at point from bot sides, and here the do not. To be captain obvious for a moment, Negative infinity does not = positive infinity. This is similar to the reason why corner points are non-differentiable as well.

Meredith G-M:
If there is a point of discontinuity on a function, the derivative does not exist at that point. This is why holes in a graph, jump discontinuities, and asymptotes are not differentiable.
Therefore if the derivative exists at a certain point, the function must be defined at that point.
The derivative of a constant function is 0. Anytime a function is horizontal at a certain point, the derivative is zero - think about local minimum and local maximum: at the points where the function levels off and becomes horizontal, the derivative is zero.

Elie P:
To answer Ms. Sweeney's question, the derivative of a constant function like f(x)=5 is zero.
Also, if the derivative exists at a certain point, we do, in fact, know that the point exists. The reason for this is that in order for the derivative to exist at a point, the function must be continuous at that point. And in order for a function to be defined at a point that point must be exist. Therefore, if the derivative exists at a point, the function is defined at that value.

Michael Joseph:
It is true that if the derivative exists at a certain point, that the function is defined at that point. It is still possible, however, for the function to be defined at a point at which the derivative does not exist. For example, if a function has a cusp x = -4, then there is an infinite amount of tangent lines to the point x =4, and therefore the derivative does not exist at this point. This represents an example of how a function can be defined when the derivative does not exist. Just because a point must exist where the derivative is defined does not mean that the derivative must exist where the function is defined.

Zach fusfeld:
Places where the function of an equation is continuous but the derivative does not exist include corner points, found on absolute value graphs, cusps, and holes (removable discontinuities). At these points the graphs are continuous. Places where the derivative does not exist due to the fact that the function is not continuous at that point include vertical asymptotes, jump discontinuites, and in many different piecewise functions



Jana: Vertical tangents, when the limits taken from both the right and left as x approaches infinity are different.
Vertical tangents occur when the derivative or f'(x) is undefined and can be seen on the graph of f.

vertical tangents and cusps _gr_5.gif]
vertical tangents and cusps _gr_5.gif]


Jana, your example is good but I think you are mixing up why a derivative doesn't exist at a vertical tangent line and the two different graphs you are given in the example that you have given. The derivative doesn't exist if we have a vertical tangent line for two reasons (that are basically the same). For starters, the slope of a vertical line is undefined, so we need to agree that the derivative (which after all IS the slope of the tangent line) is undefined. Also, even in your diagrams, if you approach the point of tangency from the left AND the right, the value of the derivative is infinity (or in the second example, negative infinity). These DO agree -- however, infinity is not a number, and since we are talking limits, it technically does not exist (for the limit to exist the function needs to approach a NUMBER).

Sample Question