Question

Given `f(x) = x^2`, considering `f(4)`, `f(5)` and `f(6)`, how do you approximate the graph of `f(x)` by a straight line at `x=5` to deduce an approximation for `sqrt(25.3)` ?

Answer 1

`sqrt(25.3) ~~ 5.03`

Explanation:

Let:

`f(x) = x^2`

Note that:

`{ (f(4) = 16), (f(5) = 25), (f(6) = 36) :}`

If we differentiate `f(x)`, then we find:

`f'(x) = 2x`

So:

`f'(5) = 10`

Alternatively, if we approximate the slope at `(5, f(5))` by the slope of a secant through the points `(4, f(4)) = (4, 16)` and `(6, f(6)) = (6, 36)`, then we find:

`m = (Delta y)/(Delta x) = (color(blue)(36)-color(blue)(16))/(color(blue)(6)-color(blue)(4)) = 20/2 = 10`

By either method, we find a value `10` for the instantaneous slope at `5`. Note that if the curve was more complex than a parabola, then these values would normally differ.

To find the (principal) square root of `25.3` we want to find `x` such that `f(x) = 25.3`, i.e. `(x, 25.3)` is a point on our parabola.

Approximating the parabola near `(5, 25)` by a straight line of slope `10`, that gives us an approximation:

`sqrt(25.3) ~~ 5+(25.3-25)/10 = 5+0.3/10 = 5.03`