Question

How do you use linear approximation to the square root function to estimate square roots `sqrt 8.95`?

Answer 1

The linear approximation of `f` at `a` is `L(x) = f(a) + f'(a)(x-a)`. This is one way of writing the equation of the line tangent to the graph of `f` at `(a,f(a))`

Explanation:

The function we want to approximate is `f(x) = sqrtx`.

We want to estimate `f(8.95) = sqrt 8.95` so we need a value for `a` that is close to `8.95` and for which we can find `f(a)`. We can readily find `f(1)` or `f(4)`, but `f(5)` would be more challenging.

Although we can easily find `f(25)`, we'd like `a` to be as close as we can get to 8.95 and still calculate `f(a)`.

By now you may be anxious for me to say, "we'll use `a=9`". So I will. We'll use `a=9`.

So `f(a) = f(9) = 3`

For the linearization we'll need `f'(9)` as well.

`f'(x) = 1/(2sqrtx)`, so `f'(9) = 1/6`

The linearization of `f(x) = sqrtx` at `a=9` is

`L(x) = 3+1/6(x-9)`

Our estimate is

`f(8.95) ~~ L(8.95) = 3+1/6(8.95-9)`

`= 3-0.05/6 = 3- 0.008bar(3)`

`= 2.991bar6`

Round to the desired accuracy. Keep in mind that this is an estimate for the desired number `sqrt8.95` so don't claim too much accuracy.

`2.99` or perhaps even `2.991` seems reasonable. (Although `2.9917` is accurate rounded to 4 decimal places.)