Question

Find the equation of tangent to the curve `y=1/(x-3)` at `x=4`?

Answer 1

Equation of tangent is `x+y=5`

Explanation:

Equation of a tangent to curve `y=f(x)` at `(x_0,f(x_0))` is

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

Here we have `f(x)=1/(x-3)` and are seeking tangent at `x=4` i.e. `x_0=4`.

As `f(x)=1/(x-3)`, `f(4)=1/1=1` and

`f'(x)=(dy)/(dx)=-1/(x-3)^2` and `f'(4)=-1`

Hence, equation of tangent is

`y-1=-1(x-4)`

or `y-1=-x+4`

or `x+y=5`

graph{(x+y-5)(y-1/(x-3))=0 [-6.83, 13.17, -5.04, 4.96]}

Answer 2

`x+y-5=0`

Explanation:

We should know that an equation of a line can be formed by `(y-y_1)/(x-x_1)=m` , in which `m` is slope of the line and `(x_1,y_1)` is the point of intersection of the curve and the tangent line.

First , we can find the first derivative `dy/dx` which is the slope of the curve.

`y=1/(x-3)`

`dy/dx=d/dx(1/(x-3))`

`=[(x-3)d/dx(1)-1d/dx(x-3)]/(x-3)^2`

`=-1/(x-3)^2`

Then, we can find the slope of the curve at `x=4`, which is also the slope of tangent line, by subing `x=4` into `dy/dx`.

`dy/dx|_(x=4)=-1/(4-3)^2=-1/1^2=-1`

Here we get `m=-1`

As, we know that the `x` coordinate of the intersection point is `4`, we can sub `x=4` into the equation of the curve to find the respective `y` coordinate.

`y=1/(x-3)=1/(4-3)=1/1=1`

Therefore, the intersection point `=(4,1)`

Now, we have enough information to form the equation of tangent line.

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

Equation of the tangent line at `x=4`:

`(y-1)/(x-4)=-1`

`y-1=-1(x-4)`

`y-1=-x+4`

`x+y-5=0`

Here is the answer. Hope this can help you :)