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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