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BlakeBoothe
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I know that it isn't. I just want to know how I could, step by step, prove that this function is not differentiable at x=0.
Welcome to Physics Forums.BlakeBoothe said:I know that it isn't. I just want to know how I could, step by step, prove that this function is not differentiable at x=0.
It's a big deal that mathematical definitions mean what they say. And it's a big deal that one's private intuitions about the way things are is not the basis for mathematical definitions or proofs. Lawyers and legal wrangling aren't popular in our culture, so writers on mathematics rarely describe math as "legalistic", but it is. If you practice it any other way you find that subjective arguments are more futile than legalism.epenguin said:I would like to ask - is this a big deal in mathematics?
In single variable calculus, the derivative is used to define the tangent line. It isn't our intuition about what a tangent line should be that defines the derivative. A person might be able to develop a consistent system of mathematics by making a different definition of derivative. (If you express a curve parametrically and pretend a particle is moving along it, you can get the kind of tangent line you want.) However, I have yet to see a definition of derivative that is more useful than the one currently used.Presumably the inverse of any continuous function is undefined at its extrema or horizontal inflection point? However ones like this still have a tangent at all points on the curve. If you just rotate the paper this inflection point is still special in one way, but no longer in that way?
So is this exercise really in the nature of a pedantry, or is there something more profound and important (and even then, is it relevant at this stage, I imagine, of the student?)
Really? Are you sure about that?kdbnlin78 said:As LCKurtz writes, the derivative here is lim_{h->0}h^{-2/3} = lim_{h->0}/cuberoot{h^{-2}} = /lim_{h->0}(/frac{1}{\cuberoot{h^2}} -> /infty.
Why should a "line" necessarily be a "function"?PraoWolf said:I'm just getting into derivatives myself and this is a problem I'm working on but the comment above concerned me. "The tangent line is a vertical line" How can a tangent line be vertical, does that not violate the definition of function?
This is badly stated. A derivative is a function, not a line, so "being vertical" makes no sense.And if it were vertical wouldn't the derivative be undefined because it's essentially showing asymptotic behavior?? While a derivative can get very very close to being vertical
A vertical line is NOT a function but that has nothing to do with it being a "line"!, I was taught once it's vertical it's no longer a function and is undefined at that point. Maybe I'm just not far enough into full understand that, but then this is Honors Calc 1.
PraoWolf said:Alright so essentially you're saying I am right. Just the fact that a line is a line and not based on a function itself. I simply meant that if you're talking about derivative and mentioning vertical lines that led me to believe that someone was saying its possible to have a tangent line that's vertical while its a line in and of itself its not a viable answer to the derivative function since it causes discontinuity.
A cubed root function is a mathematical function that takes a number as input and returns the number that, when multiplied by itself three times, gives the original input number. It can be represented as f(x) = ∛x.
The derivative of a cubed root function is defined as the slope of the tangent line to the curve at a specific point. In other words, it represents the rate of change of the function at that point. It can be calculated using the limit definition of the derivative, which involves evaluating the function at two points that are infinitesimally close together.
The cubed root of x is not differentiable at 0 because the function is not continuous at that point. The derivative of a function can only exist at a point if the function is continuous at that point. Since the cubed root function has a discontinuity at x = 0, the derivative does not exist at that point.
No, the cubed root of x cannot be made differentiable at 0. This is because the function has a sharp point at x = 0, which cannot be smoothed out or "filled in" to make the function continuous at that point. Even if we try to redefine the function to make it continuous at x = 0, the derivative still cannot exist at that point.
The fact that the cubed root of x is not differentiable at 0 means that the function is not smooth at that point. This has implications for the behavior of the function, such as the existence of a sharp corner or cusp at x = 0. It also means that the function cannot be approximated by a linear function at that point, which is a key concept in calculus and other mathematical applications.