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

1 Answer
Jan 7, 2017

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:
enter image source here
Where the red line will be the linear approximation while the blue our original function.