Linear approxiation Please help!

Question

Find the linear approximation of the function f(x) = √16-x at a = 0 and use it to approximate the numbers √15.9 and √15.99.

Answer 1

A linear approximation from a point where you know the derivative to another point on the curve looks like...

f(x+Δx) = f(x) + f'(x)Δx If you rearrange and take the limit as Δx goes to 0, this is the definition of a derivative.

Let the function sqrt(16-x) be the function. The derivative is 1/2(16-x)^-1/2 = -1/2 (1/sqrt(16-x))

^{At x = 0, f(0) = 4 and f'(0) = -1/8}

For 15.9, we can think of this as the value of f(x) where x = .1. Therefore, we can approximate this as:

f(.1) = f(0) + f'(0) * .1 = 4 - (1/8) .1 = 3..9875 which, when squared yields 15.90016

Doing the same thing for 15.99, we obtain x = 3.99875 (x² = 15.99000156)

Hope that helps.

You can also do this like a related rate or error analysis

y = sqrt(16.x) dy = -1/2(16-x)^-1/2dx Δy = -1/2(16)^-1/2Δx and y + Δy = y(corrected)