Question

How do you find the Linear Approximation at x=0 of `y=sqrt(3+3x)`?

Answer 1

I got: `y=sqrt(3)/2x+sqrt(3)`

Explanation:

The Linear Approximation should be a line that can substitute (in a narrow interval) your curve.
To find the equation of this line we need to find the slope `m` in the specific point of coordinates `(x_0,y_0)` and use the general expression for a line as:

`y-y_0=m(x-x_0)`

The slope can be found deriving our function and evaluating the derivative at `x=0`;
`y'=1/(2sqrt(3+3x))*3=3/(2sqrt(3+3x)`

At `x=0`

`y'(0)=3/(2sqrt(3))=3/(2sqrt(3))*(sqrt(3))/(sqrt(3))=sqrt(3)/2`

This will be the slope `m` of our line approximating the original curve at `x=0`.
The coordinate `y` of the specific point can be found substituting `x=0` into our original function writing:

`y(0)=sqrt(3+0)=sqrt(3)`

So, the line through our point and having slope `m` will then be:

`y-sqrt(3)=sqrt(3)/2(x-0)`
Or
`y=sqrt(3)/2x+sqrt(3)`

Graphically:

Where the red line will be the linear approximation while the blue our original function.