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 $x \geq 0$ :

$\begin{matrix} f (x) & = x (x - x) = 0 for x < 0 \\ = x (x + x) = 2 x^{2} for x \geq 0 . \end{matrix}$

Its graph is represented below:

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

\begin{matrix} f' (x) & = 0 for x < 0 \\ = 4 x for x \geq 0 \end{matrix}

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:

$\begin{matrix} f " (x) & = 0 for x < 0 \\ = 4 for x > 0 . \end{matrix}$

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 $x = 0$ .