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?

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Answer 1 · 2018-05-12T14:22:52

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:

$1/0.203 = 4.92610837438...$