Question

Let f(x) = 1/x and find the equation of the tangent line to f(x) at a "nice" point near 0.203, how do you use this to approximate 1/0.203?

Answer 1

Let `f(x) = 1/x` and evaluate its first derivative:

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

The equation of the tangent line to the graph of `f(x)` in the point `(x_0,f(x_0))` is:

`y = f(x_0) +f'(x_0)(x-x_0)`

`y = 1/x_0 -(x-x_0)/x_0^2`

Choosing:

`x_0 = 0.2 = 1/5`

the tangent line is:

`y = 5-25(x-1/5)`

and the approximation of `f(x)` for `barx= 0.203 = 1/5+3/1000` is:

`f(barx) ~= 5-75/1000 = 4925/1000 = 197/40 =4.925`

and in fact:

`1/0.203 = 4.92610837438...`