Copyright © University of Cambridge. All rights reserved.

## 'Slide' printed from http://nrich.maths.org/

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 $x\geq
0$:

$$\eqalign{ f(x) &= x(x-x)=0\ for \ x < 0 \cr &=x(x+x)=
2x^2 \ for \ x\geq 0.}$$

Its graph is represented below:

Now, I find the first derivative of $f(x)$:

$$\eqalign{ f'(x) &= 0\ {\rm for}\ x < 0 \cr &= 4x
\ {\rm for}\ x\geq 0}$$

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:

$$\eqalign{ f''(x) &= 0\ for\ x < 0 \cr &= 4\ for \
x> 0.}$$

Hence the second derivative does not exist at the origin
because on the left the limiting value of $f''(x)$ as $x\to 0$ is
$0$ whereas on the right the limiting value of $f''(x)$ as $x\to 0$
is $4$. So there isn't a unique tangent to the graph of $f'(x)$ at
$x = 0$.