Question

If this is true...
[If a function is differentiable at x=a then f in continous at x=a.]
Then why is this false?
[If a function f is continous at x=a then f is differentiable at x=a.]

Answer 1

If a function f is continous at x=a then f is differentiable at x=a

is false

look at the absolute value function

Answer 2

it's continuous at x = 0 for f(x) = |x|

but it's not differentiable there

Answer 3

Then how is the first one true then?.. it's just opposite?

Answer 4

If a function is differentiable at x=a then f in continuous at x=a

is true because the whole notion of differentiability is based on limits and continuity

Answer 5

So, if its differentiable then it's continuous. But, if its continuous then it's not differentiable.

Answer 6

correct

Answer 7

first one is always guaranteed to be true, the second isn't always true (so it's false in general)

Answer 8

haha what in the world....

Answer 9

and you can show the second isn't always true using the absolute value function as an example

Answer 10

hmm what do you mean?

Answer 11

They are not logically equivalent statements. If A is true, then B is true does not imply if B is true, then A is true. A simple example:
If a car goes 70km/h then it receives a speeding ticket.
If a car receives a speeding ticket, it is going 70km/h.

If a function is differentiable, it is continuous.
If a function is continuous it may or may not be differentiable.

Answer 12

^ that makes a lot of sense!

Answer 13

I was think that if the first statement was true, then the second had to be true.. But, I guess not.

Answer 14

:) yeah it took me awhile to sort out the formal logic they throw at you in Calculus...contrapositives and all that

Answer 15

the reason that if a function is differentiable at \(a\) then it must be continuous at \(a\) comes from the the following:
if \(f\) is differentiable at \(a\) it means
\[\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\] exists, i.e the limit exists

Answer 16

now in the above limit the denominator is going to zero
this means in order for the limit to exist, the numerator must go to zero as well, otherwise the limit would be undefined

Answer 17

ohhhh ok^

Answer 18

therefore
\[\lim_{h\to 0}f(a+h)-f(a)=0\] or
\[\lim_{h\to 0}f(a+h)=f(a)\] therfore\(f\) is continuous at \(a\)

Answer 19

as @jim_thompson5910 said, the converse if false

Answer 20

thank you (:

Answer 21

hmmmmm. @satellite73 But couldn't we argue the following?
If a function is continuous at a, then:\[\lim_{h \rightarrow 0}f(a+h)=f(a)\]So,\[\lim_{h \rightarrow 0}f(a+h)-f(a)=0\]dividing both sides by h:\[\lim_{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}\]if this limit exists. But this limit does exist since f(a+h) approaches f(a) as h goes to zero (it is continuous). So, the function is differentiable if it is continuous?

Answer 22

you could argue that, but it would be wrong

Answer 23

ok. why?

Answer 24

because in general it is possible that you have \(\lim_{x\to a} g(x)=0\) and \(\lim_{x\to a}f(x)=0\) but
\[\lim_{x\to a}\frac{f(x)}{g(x)}\] does not exist

Answer 25

ah...ok. I would love to know why!

Answer 26

getting \(\frac{0}{0}\) as a limiting form does not guarantee that the limit exists
however getting \(\frac{b}{0}\) as a limiting form if \(b\neq 0\) guarantees that the limit does not exist

Answer 27

hmmmm yeah that makes sense actually. Thinking of l'Hospital's rule for indeterminate forms and yeah I remember that some couldn't be resolved...

Answer 28

the absolute value provides the canonical example
if \(f(x)=|x|\) then \(\lim_{x\to 0}|x|=0\)

Answer 29

but \(\lim_{x\to 0}\frac{|x|}{x}\) does not exist

Answer 30

piecewise functions make a mess usually

Answer 31

yeah...makes sense.

Answer 32

^ agreed, I despise piecewise functions.

Answer 33

I recall the abs value function and the kink in it at x=0. interesting stuff

Answer 34

yeah they are a pain to write also
of course in real life (if there is such a thing) almost all functions are piecewise

Answer 35

lol