Express $y=|-x^2+1|$ as a piecewise function.

Question

I'm unsure of how to start this problem. Any help would be greatly appreciated.

Answer 1

As for real $z,$$$ |z|=\begin{cases} z &\mbox{if } z\ge0 \\ -z & \mbox{ otherwise } \end{cases}$$

$$|-x^2+1|=\begin{cases} -x^2+1 &\mbox{if } -x^2+1\ge0 \iff -1\le x\le 1 \\ -(-x^2+1)=x^2-1 & \mbox{ otherwise } \end{cases}$$

Answer 2

The absolute value of any function $y = |f(x)|$ can be defined piecewise without the use of the absolute value sign:

$$y = \left|f(x)\right| = \begin{cases} f(x) & f(x) \geq 0 \\ \\ -f(x)& f(x) < 0\\ \end{cases}$$

In your case, $y = |f(x)|$ where $f(x) = -x^2 + 1$.

So you need to determine how to express, in terms of $x$, the values satisfying $f(x) \geq 0$ and $-f(x)\lt 0$.