Use the intermediate value theorem to find the value of c such that f(c)=m.
1. f(x) = x2-4x+6 on [0,3]; m=2
if f(x) = x^2 - 4x + 6
well f(0)=6
and f(3) = 9-12+6 = 3
so in between them there is an x where f(x) is anything between 3 and 6
HOWEVER 2 is not there so I will just plain have to use the original
x^2 -4 x + 6 = 2 ????
x^2 - 4 x + 4 = 0
(x-2)(x-2) = 0
x = 2