Find the linear approximation of the function f(x) = (1 - x)^1/2 at a = 0 and use it to approximate 0.9^1/2 and 0.99^1/2

Question

Find the linear approximation of the function f(x) = (1 - x)^1/2 at a = 0 and use it to approximate 0.9^1/2 and 0.99^1/2 

 

I know how to figure this out when I'm allowed to graph it, however, my class doesn't allow me to use a calculator on the test. How do I do this?

 

#5, p255

ANSWER:

√(1 - x) ≈ 1 - x/2

√(.9) ≈ .95

√(.99) ≈ .995

Answer 1

The linear approximation function based at 0 is defined by the equation of the tangent line at 0.

 

The slope of the tangent line at 0:  f'(0) = 1/2(1-0)^(-1/2)(-1) = -1/2.

 

The point of tangency:  (0,f(0)) = (0,1).

 

The equation of the tangent line:  y - 1 = -1/2(x - 0) or equivalently y = 1 - (1/2)x.

 

The linear approximation function is y and is usually renamed L:  L(x) = 1 - (1/2)x.

 

The closer x is to the base point 0 the better L will approximate f.