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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