Question

How do you find the linearization at a=1 of `f(x) =sqrt(x+3)`?

Answer 1

Use the formula `L(x)=f(a)+f'(a)(x-a)` with `a=1` to get `f(x) approx L(x)=2+(1/4)(x-1)=x/4 + 7/4`.

Explanation:

The linearization of a differentiable function `f` at a point `x=a` is the linear function `L(x)=f(a)+f'(a)(x-a)`, whose graph is the tangent line to the graph of `f` at the point `(a,f(a))`. When `x approx a`, we get the approximation `f(x) approx L(x)`.

For `f(x)=sqrt(x+3)=(x+3)^{1/2}` we get `f'(x)=1/2 * (x+3)^(-1/2}` so that `f(1)=sqrt(4)=2` and `f'(1)=1/2 * 4^{-1/2) = 1/2 * 1/2 = 1/4`. Hence, `f(x) approx L(x)=2+(1/4)(x-1)=x/4 + 7/4`.