How to solve a twice differentiable function

Question

I assumed the functions to be quadratic and then solved for it. Image of question

I got the answer as -10

Answer 1

You can solve it as follows: $f'' = g''$ implies $f'-g' = c$ for some constant $c$, and using $f'(1) = 2$, $g'(1) = 4$ implies $c = -2$. Then $f'-g' = -2$ implies $f-g = -2x+d$ for some constant $d$, and using $f(2)=3$, $g(2)=6$ implies $d = -2$.

Hence $f-g = -2x-2$ and so $f(4)-g(4) = -2\cdot 4 -2 = -10$ as required. This agrees with you answer, without assuming the functions to be quadratic.