I want to turn below function from multipart function to a one part function ,like example: example $$f(x)=\begin{cases}3x+1 & x\geq0\\x+1 & x < 0\end{cases} \\\to f(x)=2x+|x|+1$$ now my goal is to turn $$f(x)=\begin{cases}x^2+x-2& x\leq -3\\4-3x-x^2& -3< x <1\\x^2+3x-4 & x\geq 1\end{cases}$$ I get stuck on this problem . can anybody help me ?
2026-04-06 08:07:28.1775462848
How to make one part function from this function?
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As you function brrak at $x=-3,+1 $ so must have $|(x-1)(x+3)|$ in it . so ,we can assume that $$f(x)=|(x-1)(x+3)|+g(x)$$ now look for break points $$x\geq 1 \to f(x)=x^2+2x-3+g(x)\\-3\leq x\leq 1 \to f(x)=-x^2-2x+3+g(x)$$ now compare with the original function
$$x\geq 1 \to f(x)=x^2+2x-3+g(x)=x^2+x-2 \implies g(x)=x-1\\-3\leq x\leq 1 \to f(x)=-x^2-2x+3+g(x)=-x^2-3x+4 \implies g(x)=-x+1$$ those implies $$g(x)=|x-1|$$ so merge three part of function in one part as below:$$f(x)=f(x)=\begin{cases}x^2+x-2& x\leq -3\\4-3x-x^2& -3< x <1\\x^2+3x-4 & x\geq 1\end{cases}\\=|(x-1)(x+3)|+|x-1|$$