Hello!
The formulas used to define `f(x)` are elementary functions and they are continuous on a given intervals. The only problem problematic points are the joint points `1` and `4.`
Even at these points `f` has limits from the left and from the right. And for `f` to be continuous at `1` and `4` it is necessary and sufficient for these limits to coincide. It is evident that
`lim_(x->1-) f(x) = -3,` `lim_(x->1+) f(x) = a+b,`
`lim_(x->4-) f(x) = 4a+b,` `lim_(x->4+) f(x) = 6.`
This gives the linear system for `a` and `b,` `a+b=-3` and `4a+b=6.` Solve it by substitution: `b=-3-a` and `4a-3-a=6,` thus `a=3` and `b=-6.` This is the (unique) answer.
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