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

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

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Hint: How is the absolute value function defined? – JavaMan Jun 16 '13 at 18:42

As for real $z,$$|z|=\begin{cases} z &\mbox{if } z\ge0 \\ -z & \mbox{ otherwise } \end{cases}$$ $$|-x^2+1|=\begin{cases} -x^2+1 &\mbox{if } -x^2+1\ge0 \iff -1\le x\le 1 \\ -(-x^2+1)=x^2-1 & \mbox{ otherwise } \end{cases}$$ - Why provide the whole answer for such a simple problem? Clearly the OP needs to think about these things for a while before getting them. – JavaMan Jun 16 '13 at 18:46 @JavaMan, simplicity is pretty relative, also the question is not marked 'Homework' – lab bhattacharjee Jun 16 '13 at 18:47 So you think the student will learn more by being told the answer? Regardless of not being marked as homework, it is clear that the best pedagogical practice would be to guide the student through the problem. – JavaMan Jun 16 '13 at 18:48 @JavaMan, It's not bad to have a look at solved problems solved in a standard way . Also, you will find complete answer of even simpler problems here. where you think I could leave, btw? – lab bhattacharjee Jun 16 '13 at 18:52 It's not helpful or responsible in most cases, either. You could easily have defined the absolute value, and then asked the OP where the function$-x^2 + 1$was greater than or equal to zero, and less than or equal to zero. Since our conversation has turned towards pedagogy, we should stop our discussion here. – JavaMan Jun 16 '13 at 18:55 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\$.

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