How do you find the linear approximation of `g(x)=sqrt(1+x)`?

A linear approximation of a function is a tangent line. (I can't tell you how much this realization helped me as a calculus student. It all became clear)

So we need to "linearize" at some value, call is `a`. (We find the line tangent at some point.)
We can then use the linearization (the tangent line) to approximate the function for `x` near `a`.

If you look at a textbook, you'll see that the linearization of `g` at `a` is;

`L(x)= g(a) + g'(a)*(x-a)`

Note: The equation of the line tangent to the graph of `g(x)` at `x=a` Is the equation of the line through the point `(a, f(a))` with slope `m=g'(a)`

That line, in point slope form is:
`y-g(a)=g'(a)*(x-a)`. Solve for `y` and compare to `L(x)`

Back to the question you asked about:

`g(x)=sqrt(1+x)`
So, `g'(x)=1/(2sqrt(1+x))` And `g'(a)=1/(2sqrt(1+a))`

So, the linear approximation at (or 'near') `a` is:

`L(x)=sqrt(1+a)+1/(2sqrt(1+a))(x-a)`

For a couple of choices of `a`:
`a=0`
For `x` near `0`, `g(x)=sqrt(1+x)` can be approximated by:
`L(x)=sqrt(1+0)+1/(2sqrt(1+0))(x-0)=1+1/2(x)`

`a=3`
For `x` near `3`, `g(x)=sqrt(1+x)` can be approximated by:
`L(x)=sqrt(1+3)+1/(2sqrt(1+3))(x-3)=2+1/4(x-3)`

`a=10`
For `x` near `10`, `g(x)=sqrt(1+x)` can be approximated by:
`L(x)=sqrt(1+10)+1/(2sqrt(1+10))(x-10)=sqrt11+1/(2sqrt11) (x-10)`