Slide

Age 16 to 18

Challenge Level Yellow star

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

$x = 0$ .

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