Answer

Solution

1. 1

   $f(x)=\begin{cases}x^2,& |x|\ge 1\\ |x|,& |x|<1\end{cases}$, since $|x|\ge x^2\iff |x|\le 1$.

2. 2

   $f$ is continuous (both branches agree at $|x|=1$: $|x|=1=x^2$).

3. 3

   Differentiability fails at $x=0$ (corner of $|x|$) and at $x=\pm 1$:

4. 4

   \begin{itemize}[leftmargin=*,nosep]

5. 5
   • At $x=1$: left derivative of $|x|$ is $1$; right derivative of $x^2$ is $2$. Not equal.
6. 6

   \end{itemize}

7. 7

   So $f$ is continuous but not differentiable.

Answer structure / marking notes

Content note