Question

What is the equation of the tangent line of `f(x) =sqrt((x^2-1)/(x-5))` at `x = 1`?

Answer 1

`x = 1`.

Explanation:

We start by finding the y-coordinate of the point of contact.

`f(1) = sqrt((1^2 - 1)/(1 - 5))`

`f(1) = sqrt(0/5)`

`f(1) = 0`

We now differentiate the function using the quotient and chain rules.

We call `u = (x^2 - 1)/(x - 5)`.

`u' = (2x(x- 5) - 1(x^2- 1))/(x- 5)^2`

`u' = (2x^2 - 10x - x^2 + 1)/(x - 5)^2`

`u' = (x^2 - 10x + 1)/(x- 5)^2`

We now let `y = sqrt(u)` and `u` being as noted above.

The derivative is given by `dy/dx= dy/(du) xx (du)/dx`.

`dy/dx = 1/(2u^(1/2)) xx (x^2 - 10x + 1)/(x - 5)^2`

`dy/dx = (x^2 - 10x + 1)/(2((x^2 - 1)/(x - 5))^(1/2)(x- 5)^2)`

The derivative represents the instantaneous rate of change, or slope , at any given point `x = a`.

We want to find the slope at `x = 1`.

`m_"tangent" = (1^2- 10(1) + 1)/(2((1^2 -1)/(1 - 5))^(1/2)(1 - 5)^2)`

`m_"tangent" = (-8)/(0)`

`m_"tangent" = O/`

Since the slope is undefined, the tangent is vertical. Therefore, the equation is of the form `x =a`, where `(a, b)` is the point of contact.

Hence, the equation is `x= 1`.

Hopefully this helps!