Question

How do you find the linearization of `sqrt(7+2x)` at the point a=0?

Answer 1

The answer is `=sqrt7(1+1/7x-1/98x^2+1/686x^3+o(x^3))`

Explanation:

We need

`(1+x)^n=1+n/(1!)x+(n(n-1))/(2!)x^2+(n(n-1)(n-2))/(3!)x^3+o(x^3)`

Therefore,

`sqrt(7+2x)=sqrt7(1+2/7x)^(1/2)`

`=sqrt7(1+1/2*2/7x+((1/2)(-1/2))/2*(2/7)^2x^2+((1/2(-1/2)(-3/2)))/(6)*(2/7x)^3+o(x^3))`

`=sqrt7(1+1/7x-1/98x^2+1/686x^3+o(x^3))`