1. ## Approximation and error

f(x)= x^2 -2x +3

a tangent at x=2 is y=2x=1

The tangent line is used to approximate values of f(x). A satisfactory approximation occurs if the resulting error is at most 0.4

a) Find the error when x=2.5

I won't bother showing my working for this question but my answer was 0.25

b) Show that the greatest value of x which can be used for this approximation is
(10+√10)/5

For this part though, I am just not sure what to do. If someone could show some working out or give some assistance, it would be greatly appreciated

2. ## Re: Approximation and error

let's call the line we're using to approximate f(x), g(x).

so g(x) = 2x-1.

then error = E = |f(x) - g(x)|

now f(x) - g(x) = x^2 - 2x + 3 - 2x + 1 = x^2 - 4x + 4 = (x - 2)^2

we want E ≤ 0.4, that is:

-0.4 ≤ (x - 2)^2 ≤ 0.4

since -0.4 < 0 < (x - 2)^2, for any x, we only need to check for:

(x - 2)^2 ≤ 0.4 = 4/10. taking square roots, we have:

x - 2 ≤ 2/√10

x ≤ 2 + 2/√10 = 2 + 2√10/10 = 2 + √10/5 = (10 + √10)/5

you might ask yourself...what is the smallest x could be? hint: it's NOT -(10 + √10)/5