Okay. So I am given a graph of a derivative. From what I can gather, it looks like the function might be abs(x-2)-4. (I was not given an explicit function for g', just its graph.) The question then goes on to ask me:
Is it possible, impossible, or certain that the function g is continuous at x = 2? Explain.
Is it possible, impossible, or certain that the function g is differentiable at x = 2? Explain.
I am fairly sure that the answer to the second question is certain, since the graph of g' proves that at 2, the slope of the tangent line is -4. But, I am trying to figure out what kind of function would produce a corner in its derivative. Any help is appreciated!
In that case, the derivative is continuous. The 2nd derivative will not be continuous at the cusp. The function whose derivative is the v-shaped graph, will also be continuous, but display a sharp corner where the two parabolas intersect.
|x-2| for x >/= 2 is x-2
|x-2| for x </= 2 is -x+2
that function is continuous for x = any real number
however
for x>/=2 f' = 1
for x</=2 f' = -1
so your very own function has a "corner" in its derivative at x = 2
Alright, I think I'm getting it more now. So, the function is in fact continuous at all points? But is it also differentiable at x = 2? Since I am given the graph of the derivative and x = 2 does produce a value, would it be considered differentiable?
I also know that if I can prove that the original function g is differentiable, then it must be continuous by default.
Your function has two different derivatives at x = 2. I would not call it differentiable there.
Okay, perhaps I shouldn't have given a function to work with. This proble does not given me a function, just a graph of a derivative. This graph has two straight sections, one going downward from the top left and the other increasing toward the top right. These segments meet to form a corner at (2, -4). Does this change anything?
If they form a corner, there are two values of the derivative at that point, the different slopes on each side of the corner. By the way, the second derivative is undefined there since the first derivative changes instantly.
Perhaps this from Wiki will help:
"The absolute value function is continuous, but fails to be differentiable at x = 0 since the tangent slopes do not approach the same value from the left as they do from the right."
That is from:
http://en.wikipedia.org/wiki/Derivative#Definition_via_difference_quotients
look for the graph that looks like a V
I think that I may have confused you about the graph that I am dealing with. I know that if a corner is present in an original function, then it will not be differentiable at that point. I thoroughly understand this point.
Now, the graph that I am given is the graph of the derivative. The derivative's graph has the corner in it, not the original function. So, the derivative's graph looks like a V, not the original function.
To determine whether the function g is continuous at x = 2, we need to inspect the graph of g' around that point. A function is continuous at x = 2 if and only if the graph of its derivative, g', has no vertical asymptote, jump, or point discontinuity at that point.
Looking at the graph of g', we can see that there is no vertical asymptote or jump around x = 2. However, there does appear to be a corner or cusp at that point. We can observe this by noticing that the slope of the tangent line (given by g') changes abruptly as x approaches 2 from the left and the right.
So, based on the graph of the derivative, it is impossible for the function g to be continuous at x = 2 because there is a point of discontinuity in the derivative graph.
Now let's discuss the differentiability of g at x = 2. For a function to be differentiable at a point, it must be continuous at that point as well. Since we have determined that g is not continuous at x = 2, it follows that g is also not differentiable at x = 2.
Therefore, it is impossible for the function g to be differentiable at x = 2 because it is not continuous at that point.
Regarding your question about the function that would produce a corner in its derivative, typically functions that create corners or cusps in their derivative are those involving absolute value or piecewise-defined functions. These types of functions can cause abrupt changes in the slope of the tangent line, leading to corners in the derivative graph. In this case, abs(x-2)-4 is a good example of a function with a corner in its derivative.