Exercice 1 (Fonctions dérivables) [00699]


Uploaded by Exo7Math on 09.06.2011

Transcript:
Mathematics exercises
Determine the function a, b belongs to R
so that the function f defined on R+ by:
f(x)= sqrt of x if 0 ≤ x ≤ 1
and f(x)= ax^2+bx+1 if x>1
is differentiable on R+
The function f is defined by sqrt root of x
over the interval 0 ≤ x ≤ 1
and by ax^2+bx+1 for x > 1.
We want to find a and b
so that the function f is differentiable
for all the purely positive real numbers R.
First of all,
f is clearly continuous and differentiable.
Over the open interval ]0,1[
and over the interval ]1,∞[
since it is defined by sqrt of x
which is differentiable for all the x
between 0 and 1 purely.
Pay attention, sqrt of x is not differentiable at 0
since it corresponds to a vertical tangent.
The function ax^2+bx+1
is well differentiable everywhere
and so for all x greater than 1.
The point of the exercise
is to extend the blue graph of sqrt of x
with a branch of a parabola,
the suitable branch,
so that the function becomes differentiable everywhere.
Before focusing on the differentiability,
let's look at the continuity.
We look at the limit on the left at 1.
For x<1, the function f is defined by sqrt of x.
So, when x tends to 1 by a value below,
sqrt of x tends to 1.
Now if we look at the limit on the right,
then f is defined by the formula ax^2+bx+1.
Now when x tends to 1 by a greater value,
it tends toward a+b+1.
We want the function to be differentiable,
so particularly, it has to be continuous.
So if we want f to be continuous at 1,
then it is if, and only if, we have 1+a+b=1,
in other words, the function is continuous at 1
if, and only if, we have a+b=0.
With the condition a+b=0,
the branch of the parabola has been linked together
in a continuous way with the graph
of the function of sqrt of x.
Now, we need to check the differentiability at this point.
The differentiability at 1 means
checking that the growth rate f(x)-f(1)/x-1
has a limit when x tends to 1.
Then we check the limit on the left of this growth rate,
(so the derivative on the left)
and we also check the derivative on the right.
These two derivatives will have to coincide.
The derivative on the left corresponds
to the growth rate limit for f(x) when x<1.
So we calculate sqrt of x-f(1)
which equals to 1/x-1
To calculate the limit, we are going to multiply
the numerator and the denominator with sqrt of x+1.
We notice that we have a form of (a-b).(a+b),
which gives a^2-b^2,
therefore we have x-1 in the numerator
which simplifies with the denominator
and in fact we obtain 1/sqrt of x+1.
Now, when x tends to 1, then it tends to 1/2.
Another way to obtain this value
would have been to calculate
the derivative of sqrt of x:
1/2 sqrt of x, and to calculate it with x=1
and we would have obtained the same result: 1/2.
Let's do the same on the right.
On the right, f(x) is defined by ax^2+bx+1,
therefore the growth rate becomes ax^2+bx+1
minus the value at 1 which equals to 1/x-1.
The 1 disappear; we only use a+b=0, so b=-a.
This is written in the form of ax^2-ax/ x-1
and it is simplified at ax.
So now, when x tends to 1, it tends to a.
So the function f is differentiable if, and only if,
the derivative on the right and on the left coincide
if, and only if, a=1/2.
Pay attention, with also the condition that a+b=0.
Therefore we deduct that, as a+b=0, b=-1/2.
The piece of parabola that suits
is the one for which a=1/2 and b=-1/2
and so it is the parabola graph lined here in red.
Now if we follow the blue line with the finger
and then the red line, this curve is smooth,
which means that the transition to level 1
cannot be noticed.
The derivative graphical interpretation
is the notion of tangent.
We take a point x0, f(x0)
and another one rather close, x(x),
represented here on the graph,
the point x(0), f(x0) in the graph and another point x, f(x).
If we take the straight line
that goes through these two points
(it is called a secant),
when x will tend to x0,
this secant will tend to a straight line
which will be, by definition, the tangent to the graph.
What is the slope of this dotted secant?
The slope is f(x)-f(x0)/x-x0.
If the function is derivative at x0,
this slope is the growth rate of f at x0.
Therefore it is going to tend to f'(x0).
This gives us the tangent equation at the point x0, f(x0):
it is the straight line going through x0, f(x0)
with a slope of f'(x0).
Thus we are able to find its equation.
Its equation is y=(x-x0) f'(x0)+f(x0).
Authors: Arnaud Bodin, Léa Blanc-Centi,
University of Lille 1/ Projetc 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).