Find the linear approximation to the function

Question

Find the linear approximation to the function f(x) = sqrt{x^{2}-3} at x=2 . Then, use it to approximate sqrt{6} and sqrt{13} . Without doing calculator, which of the two do we expect to have bigger error and why?

Answer 1

The linear approximation to the function is the same as the equation of the tangent line at that value of "x":

f(x) = √(x² - 3)

Therefore when x = 2: f(2) = 1

We can re-write the function to make it easier to takt the derivative using the power rule:

f(x) = (x² - 3)^1/2

f '(x) = (1/2)(x² - 3)^-1/2•(2x) = x/√(x² - 3)

Therefore f '(2) = 2

The linear approximation is y - 1 = 2(x - 2) or y = 2x - 3

To find an approximation for √6 using this function f(x) = √(x² - 3), then x² - 3 must equal 6 or x = 3

Therefore, plugging in x = 3 into y = 2x - 3 we get:

y = 2(3) - 3 = 3 so the approximate value of √6 is 3

To find an approximation for √13 using this function f(x) = √(x² - 3), then x² - 3 must equal 13 or x = 4

Therefore, plugging in x = 4 into y = 2x - 3 we get:

y = 2(4) - 3 = 5 so the approximate value of √13 is 5

The farther away you are from x = 2, the more potential error you have in using the linear approximation so √13 would likely have more error. The reason is that the function curves while the linear approximation is a straight line so the farther away you are from the point, the more likely the curve of the function is going to deviate from the line.