this post was submitted on 15 May 2024
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  • Show that if a function is differentiable for an interval, it is continuous over that interval.
  • A function is continuous if lim_x->a f(x) = f(a)
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[–] [email protected] 3 points 5 months ago* (last edited 5 months ago)

Hint:

spoiler

  • Try show that for the statement to be true, the derivative has to be defined.
  • There are 2 definitions for the derivative, either would be useful in the proof

Solution

spoilerhttps://gmtex.siri.sh/fs/1/School/Extra/Maths/Qotd%20solutions/2024-05-15_differentiablility-implies-continuity.html

[–] [email protected] 1 points 5 months ago

solutionHere's a pretty minimum effort hand-wavy proof.

A function is differentiable on an interval if it's differentiable at every point in that interval. Similarly, a function is continuous on an interval if it's continuous at every point in that interval. So I'll show that differentiability at a point implies continuity at that point, from which differentiability on an interval implying continuity on an interval will automatically follow.

A function is differentiable at x = a if lim (x → a) (f(x) - f(a)) / (x - a) exists, so let us assume it does exist.

In that limit, the denominator equals 0 at the limit point. Yet, we know by assumption that the limit exists. This implies the numerator also equals 0 at the limit point, because the only way for our denominator to equal 0 at the limit point, and still have the limit exist, is for the expression to be indeterminate, rather than undefined. The only indeterminate form with 0 in the denominator is 0/0.

So we can assume lim (x → a) (f(x) - f(a) = 0. As f(a) is a constant value, this implies lim (x → a) f(x) = f(a) - so the function is continuous at x = a.

[–] [email protected] 1 points 5 months ago (1 children)

~~Isn't the Weierstrass function a counterexample to the premise?~~ Wait no, I misread. It's the converse.

[–] [email protected] 1 points 5 months ago

i thought the "default" counter example is y=|x| lol