I wrote this on an exam, is it correct?

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In summary, the conversation discussed the concepts of continuity and differentiability, and how they are related but not interchangeable. It also touched on the definitions of continuous and differentiable functions, as well as the existence of antiderivatives for different types of functions. The question of whether the lower limit in an integral is similar to the arbitrary constant in an antiderivative was also brought up.
  • #1
1MileCrash
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F(x) always exists and is differentiable as long as f(x) is continuous.

Do you agree?
 
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  • #2
No, you have it wrong

Continuity and Differentiability
D => C but the converse C => D is not true

y = |x| is continuous at x=0 but not Differentiable there.A function is Continuous ...
Informally... if you can trace its graph without lifting your pencil
Formally ...if Limit f(x) as x-> a = f(a)

A function is Differentiable if
Informally... if it can be approximated linearly (by a tangent line) at the point in question
Formally... if the Limit of the definition exists
f ' (x) = Limit as dx -> 0 [ f(x+dx) - f(x) ] / dx

Existence and Differentiability are the same thing
 
  • #3
paulfr said:
No, you have it wrong

Continuity and Differentiability
D => C but the converse C => D is not true

y = |x| is continuous at x=0 but not Differentiable there.


A function is Continuous ...
Informally... if you can trace its graph without lifting your pencil
Formally ...if Limit f(x) as x-> a = f(a)

A function is Differentiable if
Informally... if it can be approximated linearly (by a tangent line) at the point in question
Formally... if the Limit of the definition exists
f ' (x) = Limit as dx -> 0 [ f(x+dx) - f(x) ] / dx

Existence and Differentiability are the same thing

You misunderstood his question. He was talking about primitives and integrals.

1MileCrash: yes, a continuous function [itex]f:[a,b]\rightarrow \mathbb{R}[/itex] always has a differentiable primitve function.
 
  • #4
If by F(x) you mean an anti-derivative of f(x), you are correct.
 
  • #5
Alrighty, another question on the same subject but not from the exam.

Of course the nonelementary antiderivatives inspired that question. Is the lower limit in the integral sign of a nonelementary antiderivative kind of like the + C arbitrary constant for elementary antiderivatives?
 
  • #6
1MileCrash said:
Alrighty, another question on the same subject but not from the exam.

Of course the nonelementary antiderivatives inspired that question. Is the lower limit in the integral sign of a nonelementary antiderivative kind of like the + C arbitrary constant for elementary antiderivatives?
If I understand what you're asking, there's no connection between the lower limit of integration and the arbitrary constant.

What do you mean by "nonelementary antiderivative?" I get the feeling you're really asking about definite (w. limits of integration) versus indefinite (wo limits of integration) integrals.
 
  • #7
I can tell I wasn't clear, we're running on two different terminologies, what you're referring to, we call improper integrals. By elementary antiderivative, I mean an antiderivative that is just a normal polynomial/logarithmic/what have you function.

Now that I'm on an actual PC, I can show you.

The question was:

[itex]\int 8\sqrt{\frac{3}{4} sin^{2}\theta} d\theta[/itex]

This function has no elementary antiderivative, but it does have an antiderivative. Show one, and explain why it has an antiderivative.

An antiderivative is:

[itex]\int^{\theta}_{0} \sqrt{\frac{3}{4} sin^{2}t} dt[/itex]

Because that's a function of theta increasing at the rate of:
[itex]\ 8\sqrt{\frac{3}{4} sin^{2}\theta}[/itex]

An my answer "why" was the opening of this thread.

So, my question is, since

[itex]\int^{\theta}_{0} \sqrt{\frac{3}{4} sin^{2}t} dt[/itex]

Is increasing at the same rate and therefore has the same derivative whether I replace that 0 with 6, 22, 1000, is it like an arbirary constant for a normal antiderivative, since they are all antiderivatives of the original function no matter what the lower limit of the integral is?
 

FAQ: I wrote this on an exam, is it correct?

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