Continuous at x=1,3
f(1−)=f(1)⇒ae+be−1=c...(1)
f(3)=f(3+)⇒9c=9a+6c⇒c=3a...(2)
From (1) and (2)
b=ae(3−e)...(3)
{f}^{'}(x)={\begin{matrix}a{e}^{x}-b{e}^{-x}, & -1<x<1 \\ 2cx, & 1<x<3 \\ 2ax+2c, & 3<x<4\end{matrix}
f′(0)=a−b,f′(2)=4c
Given f′(0)+f′(2)=e
a−b+4c=e...(4)
By using eq. (1),(2),(3)&(4)
a=e2−3e+13e