Local Linear Approximation?

Question

Find the local linear approximation of the function f(x)= sqrt(1+x) at x of 0 = 0.

Use the equation f(x) ~= f(x0) + f '(x0)(x-x0)

I understad the second half but I don't know how the book got 1 for f(x0) This is what I have as of now:
f(x) = f(0) + (1/2sqrt(1))(x-0)

The book has:
f(x) = 1 + (1/2sqrt(1))(x-0)

Where does the 1 come from?

Sorry if this is hard to understand. Thanks for the help!

Answer 1

A linear approximation to a function means find the best function of the form f(x) = ax + b (i.e. a straight line) which gives nearly correct values of the original function close to the given value (in this case zero).
If you put x = 0 into f(x) = sqrt(1 + x) then f(0) = 1
Are you sure that you understand why f '(0) = 1/2 ?
I am surprised that the book has the answer in that form as it would be better to just simplify it to f(x) = 1 + (x/2)

Answer 2

Just plug 0 in for the value of x in your original equation and you will get sqrt( 1+ 0) = sqrt( 1) = 1

The square root of 1 is 1.