Given points `(4, 70), (6, 69), (8, 72), (10, 81)` on the graph of a function `f(x)`, how do you find an approximate value for `f'(x)` ?

We can approximate `f'(8)` by evaluating the slope of the function between the two nearest given points `(6, 69)` and `(10, 81)`, namely:

`(81-69)/(10-6) = 12/4 = 3`

Bonus

Let's find a polynomial function that goes through these points and find out `f'(8)` for the resulting function.

Given points:

`(4, 70), (6, 69), (8, 72), (10, 81)`

Note that the `x` coordinates are evenly spaced, so let's look at the sequence of `y` values:

`color(blue)(70), 69, 72, 81`

Write down the sequence of differences between consecutive terms:

`color(blue)(-1), 3, 9`

Write down the sequence of differences between those differences:

`color(blue)(4), 6`

Write down the sequence of differences between those differences:

`color(blue)(2)`

Having arrived at a constant sequence (albeit of just one element), we can use the initial term of each of these sequences as a coefficient to give a formula for `f(x)`:

`f(x) = color(blue)(70)+(color(blue)(-1))/2(x-4) + (color(blue)(4))/(4*2)(x-4)(x-6)+(color(blue)(2))/(6*4*2)(x-4)(x-6)(x-8)`

`color(white)(f(x)) = 70-1/2x+2+1/2x^2-5x+12+1/24x^3-3/4x^2+13/3x-8`

`color(white)(f(x)) = 1/24 x^3 - 1/4 x^2 - 7/6 x + 76`

graph{1/24 x^3 - 1/4 x^2 - 7/6 x + 76 [-1, 12, 67, 83]}

Then:

`f'(x) = 1/8 x^2 - 1/2 x - 7/6`

and:

`f'(8) = 1/8 (8^2) - 1/2 (8) - 7/6 = 8-4-7/6 = 17/6`