Exercice 3 (Fonctions dérivables) [00698]

Mathematics exercises
Examine the differentiability of the following functions :
f1(x)=x^2.cos(1/x), if x other than 0 ; f1(0)=0 ;
f2(x)=sin x.sin(1/x), if x other than 0 ; f2(0)=0 ;
f3(x)=|x|sqrt (x^2-2x+1)/(x-1), if x other than 0 ; f3(1)=1.
Let f1 be defined by x^2.cos(1/x), if x other than 0 and f1(0)=0.
It is almost the same function as in the previous exercise so this will be quite easy.
First of all, according to theorems on products and compositions of functions,
the function f1 is continuous and differentiable, except possibly at 0.
Then, the limit of x^2.cos(1/x) when x tends to 0 is 0
because the cosine is limited between -1 and 1 and x^2 tends to 0.
So, by indicating f1(0)=0, we make the function f1 continuous.
Now we look at the differentiability at 0.
So we calculate the growth rate: (f1(x)-f1(0))/x-0
=(x^2.cos(1/x)-0)/(x-0)
=x.cos(1/x).
Here again, the cosine is limited between -1 and 1,
therefore when x tends to 0 the limit of this function is 0.
The growth rate has, indeed, a limit when x tends to 0 so f1 is differentiable at 0.
Here it is even possible to obtain the value of the derivative.
Same question with the function f2 which is equal to sin x.sin(1/x).
f2 is well continuous and diffrentiable except possibly at 0;
we extend it by indicating f2(0)=0 by continuity
so, it also becomes continuous at 0.
Besides we look at the differentiability at 0.
We calculate the growth rate:
(f2(x)-f2(0))/(x-0)=(sin x/x).sin(1/x).
We know that, when x tends to 0, sin x/x has a limit
and this limit is equal to 1.
However, sin(1/x) oscillates and therefore has no limit when x tends to 0.
So here the growth rate has no limit neither.
If you're not convinced, I let you write it up supposing by contradiction
that the growth rate has a limit and thus you'll show in this case that sin(1/x) has a limit
and you'll get your contradiction.
As a conclusion, the growth rate has no limit;
therefore f2 is not differentiable at 0.
Finally, the function f3 is defined by |x|sqrt (x^2-2x+1)/(x-1).
The first thing to do is to reduce the writing of f3.
Please note that (x^2-2x+1) is a binomial expansion – that is (x-1)^2 –
and the square root is the number's absolute value so we can rewrite f3 in the form of:
|x|.|x-1|/x-1.
We can obtain an expression even simpler by examining values of x.
Then, if x>1, in this case, everything which is in the absolute values is positive so :
f3=|x|.|x-1|/x-1=x
If 0≤x<1, |x|=x ; but |x-1|=-x-1.
So the quotient becomes -x.
And if x is negative, then we find x as well.
I've drawn the graph of f3 for you, which is in fact a union of affine segments.
Now according to the general theorems,
we know that – besides values of 0 and 1 –
the function f3 is continuous and differentiable.
We are interested in the point 0 and the point 1.
First of all, let's look at 0.
On the graph, 0 is obviously continuous.
Indeed, if we look at the limit on the left at 0, it tends to 0;
the limit on the right at 0 tends to 0; therefore f3 is continuous at 0.
So here we are interested in the differentiability.
We are going to calculate the growth rate on the left
When we are on the left, this means we are in values less than 0.
So the growth rate equals to:
(x-0)/(x-0)=+1, so it is equal to 1.
If we look at the growth rate on the right, in this case,
we are in values of x greater than 0 so :
f3(x)=-x/x-0=-1.
Thus, the growth rate is constant.
We can see that the growth rate on the left and the one on the right have not the same limit.
So, f3 is not differentiable at 0.
It was quite obvious since we can see that there is a tangent on the left
which is parallel to the one on the right
(which has the same slope than the straight line)
and a tangent on the right which has the same slope (-1).
Now let's look at 1:
It is even faster at 1.
We immediately see on the graph that f3 is not continuous at x=1.
This can be easily shown when looking at the limit on the left at 1
and the limit on the right at 1.
Therefore it is not differentiable neither.
To conclude:
Besides 0 and 1, f3 is continuous and differentiable;
at 0, f3 is continuous but not differentiable;
and at 1, f3 is neither continuous nor differentiable.
By definition, f is differentiable at x0
if the growth rate (f(x)-f(x0)/x-x0 tends to a finite limit, noted down l.
We can reformulate this condition of differentiability in a way equivalent to the following:
f(x) can be written in the form of :
f(x)=f(x0)+(x-x0).l+Ɛ(x).
(x-x0) or Ɛ(x) is a function that tends to 0 when x tends to x0.
With the help of this second writing, it is easy to show the following property :
If f is differentiable at the point x0, then f is continuous at x0.
And in a more general way :
If f is differentiable on an interval I, then f is continuous on I.
It should always be said that the reciprocal is wrong.
Indeed, for instance, the function that has x associated with |x|
is continuous at 0 but is not differentiable at 0.
Therefore, remember the graph:
we see very well on the graph that the function absolute value is continuous,
but if we calculate the growth rate on the left and the growth rate on the right,
they have not the same limit when x tends to 0,
and this is obvious when we look at the tangents on the left and the tangents on the right.
Authors: Arnaud Bodin, Léa Blanc-Centi, university of Lille 1/ Project Exo7.
Shooting/Editing: Guy Vantomme, SEMM-Lille 1
Production/ Technical means: SEMM, Teaching and Multimedia service
University of Lille 1
UNISCIEL (The Online Thematic University for Sciences).