Question

How do you find the points where the graph of the function `y=(4x)/(x+1)` has horizontal tangents and what is the equation?

Answer 1

Find the derivative, set it to `0`, solve the equation, then find the `y` value(s) at the solution(s). The horizontal tangent at that point has equation `y=` the value.

Explanation:

`y = (4x)/(x+1)` has derivative `y' = 4/(x+1)^2` which is never `0`, so there is no horizontal tangent line.

Finding `y'`

Use the quotient rule:

`y' = ((4)(x+1) - (4x)(1))/(x+1)^2 = (4x+4-4x)/(x+1)^2 = 4/(x+1)^2`

A different example

`y = x^3-3x^2+5`

`y' = 3x^2-6x = 0` at `x=0` and at `x=2`

At `x=0`, we get `y=5` so the line tangent to the curve at `(0,5)` has equation `y=5`.

At `x=2`, we get `y=1` so the line tangent to the curve at `(2,1)` has equation `y = 1`.