Let the number be represented as ABCDEF, where A, B, C, D, E, and F are the digits.
Given the conditions:
\(C=A\)
\(B=2A\)
\(F=A+B+C=A+2A+A=4A\)
\(E=A+B=A+2A=3A\)
\(D=E+F=3A+4A=7A\)
Since A and D are both digits, the maximum possible value of A is 1. Therefore, the maximum value of D is 7.
Define \( f(x) = \begin{cases} x^2 + bx + c, & x< 1 \\ x, & x \geq 1 \end{cases} \). If f(x) is differentiable at x=1, then b−c is equal to