Piecewise[{{(x^2-1)/(x-1),x1},{ax^2-bx+1,1=x3},{4x-a+b,x=3}}] solve for a and b

Piecewise[{{(x^2-1)/(x-1),x<1},{ax^2-bx+1,1<=x<3},{4x-a+b,x>=3}}] solve for a and b

The function is piecewise continuousFor x&lt;1, f(x) = x2 – 1 / (x – 1)f(x) = x + 1So at right boundary, f1(1) = 1+ 1 = 2For 1&lt;=x&lt;3, f(x) = ax2 – bx +1At left boundary, f2(1) = a...