f(x)=ax+b, 2a^2 + 6b^2 =3, prove |f(x)| is less than/equal to sqrt(2)

given f(x) = ax + b 

x belongs to the range [-1, 1]

and 2a^2 +6b^2 = 3

please prove that |f(x)| is less than or equal to sqrt(2)

 

I am supposed to do this with trigonometry/trigonometric identities, but I'm not sure how to do that.

Is there a way to prove this without the use of a graphing calculator, or without trig?

 

please help asap, thanks!