Question

How do you find the tangent line approximation to `f(x)=1/x` near `x=1` ?

Answer 1

`L(x) = 2-x`

We find the answer from the linear approximation `L(x) = f(a)+f'(a)(x-a)`. For linear approximations, we want both `f(a)` and `f'(a)` so that they are easy to calculate. In this case, `f(1)` and `f'(1)` are easy.

To find the derivative of

`f(x) = 1/x`

it is best to rewrite it in power form:

`f(x) = x^(-1)`

Using the power rule for derivatives:

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

So substituting, `a=1`:

`f(a)=1/1`

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

Substituting this into `L(x)`,

`L(x)=1-1(x-1)`

`=1-x+1`

`=2-x`