Evaluate the function ƒ( x ) = x 2 + 1 for ƒ(2).
Plug in the 2 for the x square it and add 1 and u'll get 5
If \(f(x) = x^{2} + 1\)then \(f(2) = (2)^{2} + 1\)∴ f(2) = (4) + 1 = 5 So in essence, whatever is the the bracket adjacent to the f, in this case 2, you substitute that for the x or x's in the equation and then solve.
