This solution was written by Andrei from Tudor Vianu National College,
Bucharest, Romania.
I calculate the values of f(x) = x(x+|x|) for x < 0 and x ³ 0:
f(x)
= x(x-x)=0 forx < 0
=x(x+x) = 2x2forx ³ 0.
Its graph is represented below:
Now, I find the first derivative of f(x):
f¢(x)
= 0 for x < 0
= 4x for x ³ 0
I observe that for x = 0, f¢(x) is 0 both from the first and
from the second form of f(x), that is on both sides of the origin.
So the first derivative f¢(0)=0 exists at x=0. Hence the graph
of f(x) has the tangent at y=0 at the origin
Now, I calculate the second derivative:
f¢¢(x)
= 0 forx < 0
= 4 forx > 0.
Hence the second derivative does not exist at the origin because on
the left the limiting value of f¢¢(x) as x® 0 is 0 whereas on
the right the limiting value of f¢¢(x) as x® 0 is 4. So there
isn't a unique tangent to the graph of f¢(x) at x = 0.
