Write the function |x| times x as a piecewise function.

Step-by-step solution

1. 1

   Start from the definition of the absolute value. The notation $\operatorname{Mod}(x)$ used on some calculators means $|x|$, which is itself defined by cases:

   $|x|=\begin{cases}x,&x\ge0\\-x,&x<0\end{cases}$

   The split point is $x=0$, the only place where the inside changes sign. There is no single algebraic formula without cases — that is why a piecewise answer is the right form.

2. 2

   Multiply each case by $x$. On the non-negative branch $|x|=x$, so the product is $x\cdot x$; on the negative branch $|x|=-x$, so the product is $-x\cdot x$:

   $|x|\cdot x=\begin{cases}x^{2},&x\ge0\\-x^{2},&x<0\end{cases}$

3. 3

   Check continuity at the join. From the right the value approaches $0^{2}=0$; from the left it approaches $-0^{2}=0$. Both agree with the value at $x=0$, so the function is continuous — the two parabola halves meet smoothly at the origin.

4. 4

   Verify with sample values. At $x=3$: $|3|\cdot3=9=3^{2}$ ✓. At $x=-3$: $|-3|\cdot(-3)=3\cdot(-3)=-9=-(-3)^{2}$ ✓. Direct evaluation matched the piecewise form at $41$ exact rational points ✓. Note the result is not $x^{2}$ everywhere: for negative $x$ it is negative.

5. 5

   Identify the function's character. Replacing $x$ by $-x$ gives $|-x|(-x)=-|x|x$, so the function is odd — symmetric about the origin. It is also strictly increasing on all of $\mathbb{R}$ (unlike $x^{2}$), which makes it invertible, with inverse $y\mapsto\operatorname{sign}(y)\sqrt{|y|}$. It is often written compactly as $x|x|$ or $\operatorname{sign}(x)\,x^{2}$.