The linear approximation to (1/sqrt(9-x)) at x=0. What is y?

Answer 1

the linear approximation is just the tangent line at the given point. Within a suitably small interval, the line approximates the curve to any desired accuracy.

So,

y = 1/√(9-x)
y' = 1 / 2(9-x)^(3/2)
at (0,1/3), y' = 1 / 6√3

So, now we have a point and a slope. The line is

y - 1/3 = (1/6√3)(x-0)