Question

Why is 1/x not continuous and yet is differentiable?

It mentions at the end of the solution in the worked example for session 2 that a function which can not be differentiated at even one point in it's domain it is not a differentiable function. For 1/x it can not be differentiated at x=0, but the function is differentiated in the lecture. It should be dy/dx=-1/x^2.

If someone could shed some light on this I would be grateful. Thanks.

Answer 1

I'd say that 1/x is continuous and differentiable everywhere it is defined, ie\[]-\infty;0[u]0;+\infty[\]
f(x)= |x|, is defined and continuous in 0, yet not differentiable - that is the difference

Answer 2

Thankyou for the response. I've not come across that notation before and am unsure of what the 0[u]0 means. Maybe I've inferred too much about what discontinuity means too. I thought that 1/x had (as defined in the session 5 lecture) infinite discontinuity.

Perhaps it is a special case and I missed something. Why can it have two different derivatives at x=0 (positive and negative infinity) and still be classified as well defined?

Thanks.

Answer 3

Sorry about the notation, it should have been U not u - anyway it means that f(x)'s DOMAIN is composed of two Intervals :from minus infinity to 0 excluded, ie 'just before' 0, and from 'just after' 0 to + infinity - or more simply put, R - {0}. The function 1/x 'exists' everywhere except for 0.
The sentence that's causing the confusion is:
"If a function f(x) is not differentiable at even one point in its domain, f(x) is not a differentiable function."
Since f(x)=1/x isn't DEFINED for x=0 (ie, 0 is not in it's domain), this sentence does not apply. And of course you cannot differentiate a function outside of its domain.

You're right, depending on where you're coming from 1/x goes to + or - infinity as x TENDS to 0 (depending on whether you're coming from the right, ie 0+ or the left, ie 0-). The key word here is tends, which means that x gets closer and closer to 0 (infinitely close) but NEVER hits 0.

This is no special case - it's true each time 'division by 0' occurs.: y will tend towards + or - infinity (depending on whether the numerator is positive or negative) -it's called a vertical asymptote.

btw a function isn't 'well' defined - it just is or isn't!

I'm pretty new to all this myself and I think the key concepts here are domain and limits, as much as continuity and differentiability.

Answer 4

Excellent. This clears everything up. Thanks a lot for your help. I'm also really glad I asked it now too.