Linear approximation to approximate a value

[PhizKid]

Homework Statement

f(x) = sqrt(1 - x)
a = 0

Approximate sqrt(0.9).

Homework Equations

L(x) = f'(a)(x - a) + f(a)

The Attempt at a Solution

I understand that linear approximation is finding the equation of a line of a point tangent to a function. But now this question is asking me to approximate the value of sqrt(0.9), which I assume means without the use of a calculator. The given function is sqrt(1 - x) and a = 0. So f(0) = 1 and f'(0) = -1/2. Therefore the equation of the line tangent at (0,1) of this function is -x/2 + 1. How does this have anything to do with finding the value of sqrt(0.9)?

 

[Dick]

Homework Statement

f(x) = sqrt(1 - x)
a = 0

Approximate sqrt(0.9).

Homework Equations

L(x) = f'(a)(x - a) + f(a)

The Attempt at a Solution

I understand that linear approximation is finding the equation of a line of a point tangent to a function. But now this question is asking me to approximate the value of sqrt(0.9), which I assume means without the use of a calculator. The given function is sqrt(1 - x) and a = 0. So f(0) = 1 and f'(0) = -1/2. Therefore the equation of the line tangent at (0,1) of this function is -x/2 + 1. How does this have anything to do with finding the value of sqrt(0.9)?

The tangent line is close to the curve sqrt(1-x) if x is close to zero. So you can approximate value of the curve by the value of the tangent line. What value of x would you use?

 

[PhizKid]

Oh, 1 - x = 0.9 so 0.1

Okay, I got it now