While solving this question I ran into some trouble.
Consider using Mean value theorem we have :
$f(1)= 6$ and $f(0)=2$
Thus, $$ \frac {f(1)-f(0)}{1-0}=f'(c_1) $$ for some $ 0 \lt c_1 \lt 1$
And similarly for $g(x)$ we have
$g(1)=2$ and $g(0) = 0$
So,
$$ \frac {g(1)-g(0)}{1-0}=g'(c_2)$$
for some $ 0 \lt c_2 \lt 1$
Solving these we get $f'(c_1) = 2\ g'(c_2)$
This suggests that $c_1$ and $c_2$ are two different terms.
But now consider $h(x) =f(x) -2 g(x)$
Note that $h(1) =h(0)$ so now we can apply Rolle's theorem which gives :
$h'(c) =0 $ for some $ 0 \lt c\lt 1$
This means $h'(c) = f'(c)-2g'(c) =0$
$$\implies f'(c) =2g'(c)$$
This means that $c_1 =c_2 =c$ .
But consider $f(x) = 4x^2 +2 $ and $g(x) = 2x^3$ here solving we get $c_1 = {1 \over 2}$ and $c_2 = {1\over \sqrt 3}$
Question : Why does Rolle's theorem seem to give same value for $c$ while the above example contradicts this?
Rolle gives a point where $f'=2g'$, the other methods give a point where $f'=4$ and an unrelated point where $g'=2$. Those are two ver different tasks.
Sayin that there is a woman Jill making 400000\$ a year and there exists a man John making 200000\$ a year is a completely different statement than saying that there exists a married woman making twice as much as her husband. In particular, we cannot conclude that Jill and John are married.