Answer:
Do you want to be extremely boring?
Since the value is 2 at both 0 and 1, why not make it so the value is 2 everywhere else?
[tex]f(x) = 2[/tex] is a valid solution.
Want something more fun? Why not a parabola? [tex]f(x)= ax^2+bx+c[/tex].
At this point you have three parameters to play with, and from the fact that [tex]f(0)=2[/tex] we can already fix one of them, in particular [tex]c=2[/tex]. At this point I would recommend picking an easy value for one of the two, let's say [tex]a= 1[/tex] (or even [tex]a=-1[/tex], it will just flip everything upside down) and find out b accordingly:[tex]f(1)=2 \rightarrow 1^2+b+2=2 \rightarrow b=-1[/tex]
Our function becomes
[tex]f(x) = x^2-x+2[/tex]
Notice that it works even by switching sign in the first two terms: [tex]f(x) = -x^2+x+2[/tex]
Want something even more creative? Try playing with a cosine tweaking it's amplitude and frequency so that it's period goes to 1 and it's amplitude gets to 2: [tex]f(x) = A cos (kx)[/tex]
Since cosine is bound between -1 and 1, in order to reach the maximum at 2 we need [tex]A= 2[/tex], and at that point the first condition is guaranteed; using the second to find k we get [tex]2= 2 cos (k1) = cos k = 1 \rightarrow k = 2\pi[/tex]
[tex]f(x) = 2cos(2\pi x)[/tex]
Or how about a sine wave that oscillates around 2? with a similar reasoning you get
[tex]f(x)= 2+sin(2\pi x)[/tex]
Sky is the limit.
