\(f(x)=\begin{cases}\\4-2x-x^2 &: x<-1 \\ ax+b &: -1\le x\le 1\\ x^3 &: 1< x\end{cases}\)Solve for the constants a and b that make the piecewise function continuous for all real numbers. f(x)={\\ 4-2x-x^2, x<-1 ax+b, -1 ≤x ≤1 x^3, x>1
Sign error?Since \(\displaystyle f(x)= 4- 2x- x^2\) for x less than -1, the limit "as x approaches -1 from below" is \(\displaystyle \lim_{x\to-1} 4- 2x- x^2\) which, since \(\displaystyle 4- 2x- x^2\) is a continuous polynomial, is \(\displaystyle \huge{4- 2(-1)- (-1)^2= 4+ 2\color{red}+ 1= 7}\).
You didn't indicate why that stopped you; I'm not sure what it is.I understand that part. Normally when I do questions like this I’d find what f(-1) is and then find the limits x->1- and x->1+ but I can’t do that with this one because of ax+b, -1 ≤x ≤1. So how do I start the problem?