Derivative at a possible corner

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In summary: So the limit is not divergent.The limit definition of the derivative for this particular f(x) doesn't seem to exist.The limit just doesn't exist :-)
  • #1
Qube
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Homework Statement



http://i.minus.com/j1DgeK5VMMfHn.png

Homework Equations



To simplify a limit with a square root multiply by the conjugate.

The Attempt at a Solution



The limit doesn't seem to exist as the left and right hand limits diverge.

The left hand limit with f(x) = x^2 + 4 results in x^2/x = x and the limit of x as x approaches 0 from the left is 0.

The right hand limit, after multiplying it by its conjugate, simplifies to 1/[(sqrt(x+16)) + 4]. Taking the limit of this as x approaches 0 results in 1/8.

The limit definition of the derivative for this particular f(x) doesn't seem to exist.

Does this imply the function is not differentiable at x = 0? I graphed it out and it seems there is a corner at x = 0. I tried looking up the mathematical definition of a corner but the best I got was that at a corner the derivative changes abruptly; it's not "smooth." The derivative does jump from being negative on the left with x^2 + 4 and to being positive with the sqrt(x+16). So what's the mathematical definition of a corner? I know for a cusp the mathematical definition is that the left and right hand limits go to infinities of different signs at the point of the cusp. So what's a corner? And am I correct that this function is not differentiable at x = 0?
 
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  • #2
Qube said:
The limit doesn't seem to exist as the left and right hand limits diverge.

Be careful with the language here. The word diverge - in "f diverges as ##x \to a##", is usually used to mean the formal equivalent of "##\lim_{x \to a} f(x) = \infty##". What you mean is that the left and right hand limits are not equal, and therefore the limit does not exist. There is no divergence though.

The left hand limit with f(x) = x^2 + 4 results in x^2/x = x and the limit of x as x approaches 0 from the left is 0.

The right hand limit, after multiplying it by its conjugate, simplifies to 1/[(sqrt(x+16)) + 4]. Taking the limit of this as x approaches 0 results in 1/8.
That's right. Note that these one-sided limits are actually the derivative of the pieces evaluated at x = 0: ##d/dx(x^2 + 4)\vert_{x = 0} = 0## and ##d/dx( \sqrt{x + 16} )\vert_{x = 0} = 1/8##. Of course, you noticed that but you were asked to use the limit definition.

The limit definition of the derivative for this particular f(x) doesn't seem to exist.
The definition is fine, the limit just doesn't exist :-)

Does this imply the function is not differentiable at x = 0? I graphed it out and it seems there is a corner at x = 0. I tried looking up the mathematical definition of a corner but the best I got was that at a corner the derivative changes abruptly; it's not "smooth." The derivative does jump from being negative on the left with x^2 + 4 and to being positive with the sqrt(x+16). So what's the mathematical definition of a corner? I know for a cusp the mathematical definition is that the left and right hand limits go to infinities of different signs at the point of the cusp. So what's a corner? And am I correct that this function is not differentiable at x = 0?

I'm not sure "corner" has an official definition in the context of function graphs (probably it is defined when talking about e.g. polygons). You could define it as a point where the function is continuous but not differentiable. I don't know whether "cusp" is formally defined either, but I would expect it to be similar - in any case I don't think it's necessary for any limit to go to infinity. E.g. I would say f(x) = |x| has a cusp or corner at x = 0, but I wouldn't say that f(x) = 1/x does, while only the latter involves "infinities" - that is, one-sided limits that do not exist.
 
  • #3
Thank you for the thoughtful reply! I agree that precise wording is an imperative.

I found a definition for a cusp on the University of Chicago's website.

http://math.uchicago.edu/~vipul/teaching-1011/152/concaveinflectioncusptangentasymptote.pdf

http://i.minus.com/jZ459eGBlUzbv.png
 
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  • #4
Re cusp as per your definition not ##x^{2/3}## which has no derivatives going off to infinity at any finite point. But 1/x is an example.
 

Related to Derivative at a possible corner

What is a derivative at a possible corner?

A derivative at a possible corner is a mathematical concept that describes the rate of change of a function at a point where the function is not differentiable. In other words, it is the slope of a tangent line at a point where the function has a sharp turn or corner.

How do you find the derivative at a possible corner?

To find the derivative at a possible corner, you can use the limit definition of a derivative. This involves taking the limit of the slope of a secant line as the two points on the line get closer and closer to the point in question. If the limit exists, it is equal to the derivative at the possible corner.

What does a derivative of zero at a possible corner mean?

A derivative of zero at a possible corner means that the function is not differentiable at that point. This could be due to a sharp turn or corner in the function, or because the function is not continuous at that point. In general, a derivative of zero at a possible corner indicates a discontinuity in the slope of the function.

Can a function have a derivative at a possible corner?

Yes, a function can have a derivative at a possible corner. This occurs when the function is not differentiable at that point, but the limit of the slope of the secant line exists and is equal to the derivative. This is known as a removable discontinuity, as the derivative can still be defined at the possible corner point.

Why is it important to consider derivatives at possible corners?

Considering derivatives at possible corners allows us to better understand the behavior of a function and its rate of change. It also allows us to identify points of discontinuity and better analyze the graph of a function. In applications, derivatives at possible corners can help us model and predict real-world situations more accurately.

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