Derivative Graph

[chowe13]

I have a question. If the graph of the derivative is plotted ,and at some point c there is a short vertical segment on the derivative graph, what is happening at this point on the original function. Where it is vertical is is still always increasing both before and after the segment, though it appears it could be an inflection point on the graph of the derivative as well.

 

[renegade05]

I have a question. If the graph of the derivative is plotted ,and at some point c there is a short vertical segment on the derivative graph, what is happening at this point on the original function. Where it is vertical is is still always increasing both before and after the segment, though it appears it could be an inflection point on the graph of the derivative as well.

This would be impossible. The derivative is the slope at point c. So this would mean for a vertical segment to exist, the slope would have to be: \(\displaystyle \frac{rise}{run}\) where run=0. But you cannot divide by zero. Hence, not possible.

Also, if a graph does have a vertical segment, this is no longer a function. It fails the vertical line test.

You can however find where there are vertical segment's on a parametric graph. You can do this by solving when the change in x is zero.

 

[lookagain]

If the graph of the derivative is plotted ,and at

some point c there is a \(\displaystyle > > \)short vertical segment\(\displaystyle < < \) on the derivative graph,

what is happening at this point on the original function.

Where it is vertical is is still always increasing both before and after the segment,

though it appears it could be an inflection point on the graph of the derivative as well.

\(\displaystyle > > \) ? ? \(\displaystyle < < \)

\(\displaystyle \text{What if the original function were \ \ }\)
\(\displaystyle \ \ f(x) = x^{\frac{1}{3}} \ ?\)


\(\displaystyle \text{Notice a few of these if you have a calculator and graph it:}\)


There is a vertical tangent, x = 0, that passes through the point (0, 0).


The concavity changes from up to down on either side of the point (0, 0) ,
and the function is increasing on the interval \(\displaystyle (-\infty, \infty).\)


Then the point (0, 0) should be an inflection point.


The first derivative is \(\displaystyle \frac{1}{3}x^{\frac{-2}{3}} \ = \ \dfrac{1}{3x^\frac{2}{3}}\)


Then the slope at the point (0, 0) is undefined.


Testing x values on each side of x = 0 in the first derivative shows
that the function is increasing on both sides of x = 0.


The second derivative is \(\displaystyle \ \frac{-2}{9}x^{\frac{-5}{3}} \ = \ \dfrac{-2}{9x^{\frac{5}{3}}}\)


Testing x values on each side of x = 0 in the second derivative shows
that the function is concave up to the left of x = 0 and concave
down to the right of x = 0.


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chowe13, does my example answer any of your questions?

 

[chowe13]

I was thinking that it could be a cusp where f(x) is no longer a function, but wouldn't this mean that the original graph would not be differentiable at that point. What is weird is that the graph of teh derivative is still continuous. I looked at the cube root of x as well but I thought that it would be non-differentiable at 0, not a vertical line on the derivative (i know the original function would have a vertical tangent but also thought it would not be continuous on the derivative. The only possibilities I could come up with are:

1. cusp or rapid change such as cube root of x

2. cusp involving a non-function for the original graph, so involving implicit differentiation

But I am not sure how the derivative of either of these would even exist at those points. I thought it would be an asymptote at the point, with the graph of the derivative going to positive and negative at each side. thanks