How do can you prove that a function is both continuous and differentiable?

Question

I want a mathematical explanation...thanks in advance!

Answer 1

Ooo I know this one! :) I did really well on this part of calculus at the beginning of this school year.


The definition of continuity has 3 parts:

1) lim f(x) as x->c exists

2) f(c) exists

3) lim f(x) as x-> c equals f(c)


thats all there is to proving continuity. you can write it all out on one line for simplicity, but its hard for me to show you on this computer. you know how for a limit of f to exist, its limit from the left (-) and right (+) have to equal each other? ok, well, when i type out the line of proof that you need to write, im going to write "lim-" and "lim+" to mean "the limit of f(x) from the left/right as x->c"

to prove that f(x) is continuous at x=c . . .

lim- = lim+ = f(c)



Differentiability is similar, but with slope (because a derivative is just the slope of the tangent line). To prove that f(x) is differentiable at x=c . . .

1) prove that f(x) is continuous at x=c

2) show that the slope from the left of c equals the slope from the right of c

thats all!


I hope that was useful!

Answer 2

Correct me if I'm wrong, but I'm guessing that you're talking about the kind of problem in which you have a function defined in pieces, such as

f(x) = x+1 for x<0, and x^2+x+1 for x>=0.

This function, for example, is both continuous and differentiable at x=0.

To see this, first compute the left-hand and right-hand limits at x=0, as well as f(0). For f(0), use x^2+x+1, and you get f(0)=1.

For the left-hand limit, you want to consider x<0, so you use x+1, plug in x=0, and get 1.

For the right-hand limit, you want to consider x>0, so you use x^2+x+1, plug in x=0, and get 1.

The left-hand and right-hand limits are equal, so you can say lim (x->0) f(x)=1. Since f(0)=1, it follows from the definition of continuity that f is continuous at 1.

You can proceed in a similar fashion for differentiability, except that you work with the derivatives of both formulas and plug in x=0. That tells you the left-hand and right-hand derivatives at x=0. If they agree, then the function is differentiable at 0.

Answer 3

For a function to be differentiable at a point, the left-hand and right-hand limits must be the same. For it to be continuous, those limits must be equal to the value of the function at that point.

Are you asked to prove it's continuous and differentiable at a specific point? For a given domain? Graphing is a good way to do this.

Answer 4

Every differentiable function is automatically continuous.
If it's a function f from R to R, then you have to prove that, for every x in it's domain, there exists the limit f'(x) = lim (h->0) (f(x + h) - f(x))/h.

There´s no general answer, each function has to be considered individually.

Answer 5

No, that suffices. yet extra beneficial than in all probability the way the function is defined, you won't be able to honestly prepare that f(x) = g(x) and f'(x) = g'(x). One piece is probable defined with a strict > or < sign, and the different is probable defined with a >= or <=. which ability in basic terms between the products is honestly defined on the "factor of circulate". So the argument you're thinking of ought to be finished formally with the barriers of the two products and their derivatives as x procedures that factor from the two factors. you're showing the minimize is the comparable as you attitude from each direction. and that's what you could prepare. Do you preserve on with? If I say f(x) = x^2, x > 2, then i can not declare f(2) = 4 because of the fact f(x) isn't defined at x = 2. i can in basic terms declare the minimize of f(x) as x->2 from above is 4.

Answer 6

A function that is differentiable is continuous, the proof is in most calculus texts, so half of your problem is done.

I am rusty on this stuff, that is all I know for now.