How do you use linear approximation to the square root function to estimate square roots `sqrt 4.400`?

Using binomial expansion,

`sqrt(a^2+b)=a(1+b/a^2)^(1/2)=a(1+1/2b/a^2)`, nearly.

Here, choose a = 66 and b = 44.

Now,

`sqrt4400=sqrt(66^2+44)=sqrt(66^2+44)=66(1+44/66^2)^(1/2)`

`66(1+1/2(44/66^2))=66(1+1/198)=66+1/3=66.33`, nearly.

Note that the error is of order 66(1/198)^2=O(.002), nearly.

For function, `f`, the linear approximation at `a` is given by

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

Note that the equation of the line tangent to the graph at `(a,f(a))` has slope `m=f'(a)` and it has point-slope equation

`y-f(a) = f'(a)(x-a)` Solving for `a` gives us the linear approximation to `f` at `a`.

We are not told what to use for `a`, but we want to eventually use `x = 4.4`.

We want `a` to be a number "close to" 4.4 for which it is relatively easy to calculate `f(a)`.

In this case, we want a number close to `4.4` whose square root is relatively easy to find.

The 'obvious' (once you see it) choice is `a = 4`.

So,

`f(x) = sqrtx` and `f(a) = f(4) = 2`.

`f'(x) = 1/(2sqrttx)` so `f'(a) = f'(4) = 1/4`

`L(x) = 2 + 1/4(x-4)`.

And we finish with

`L(4.4) = 2+1/4(4.4-4) = 2.1`

If there is a typo in the question

If the question should ask us to approximate `sqrt(4,400)` we should find a different `a`.

`60^2 = 3600` and `70^2 = 4900` so, for a rough approximation, use `a=70`.

A bit more arithmetic will show that

`66^2 = 4356` and `67^2 = 4489`.

Since `4356` is close to `4400`, use `a=4356`.

OR

To get a rough estimate use `sqrt4400 = sqrt(4 xx 100 xx11) = 20sqrt11` and approximate `sqrt11` using `a=9`.