Question

How do you find the slope & equation of tangent line to `f(x)=-1/x` at (3,-1/3)?

Answer 1

Slope `=1/9` & equation: `x-9y-6=0`

Explanation:

Given function:

`f(x)=-1/x`

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

Now, the slope `m` of tangent at the given point `(3, -1/3)` to the above function:

`m=f'(3)`

`=1/3^2`

`=1/9`

Now, the equation of tangent at the point `(x_1, y_1)\equiv(3, -1/3)` & having slope `m=1/9` is given following formula

`y-y_1=m(x-x_1)`

`y-(-1/3)=1/9(x-3)`

`9y+3=x-3`

`x-9y-6=0`

Answer 2

`y=1/9x-10/3`

Explanation:

`"the slope of the tangent line is "f'(x)" at "x=3`

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

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

`f'(3)=1/(3^2)=1/9`

`y+1/3=1/9(x-3)`

`y+1/3=1/9x-1/3`

`y=1/9x-10/3`