To be continuous, the functions should have the same value at the transition values of x
First transition value: x = 0
at x = 0, 3x^2 - 2x + 1 = 1
and acosx + b = acos(0) + b = a+b
since they wanted the value of a+b and we know its value is 1
we are done: a+b = 1
at x= π/3
acosπ/3 + b = a/2 + b
and 4sin^2 (π/3) = 4(√3/2)^2 = 4(3/4) = 3
then a/2 + b = 3
a+2b = 6
solving with a+b=1, subtract them
b = 5
then a+5 = 1
a = -4
so when x=0 , first function is 1
2nd function is -4cos0 + 5 = -4+5 = 1 , good
2nd function is -4cos(π/3) + 5 = -4(1/2) + 5 = 3
3rd function is 4sin^2 (π/3) = 4(√3/2)^2 = 3 , good!
So the values of a=-4 and b=5
make the functions continuous.
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