Question

Can someone please explain to me what this question is asking and how I should go about solving it: Of all the lines tangent to the curve f(x)=6/(x^2+3) find the x-coordinate of the tangent lines of minimum and maximum slope?

Answer 1

Here is a solution that uses calculus (not a precalculus solution).

Explanation:

For `f(x)=6/(x^2+3)`, the slope of the tangent line is given by

`f'(x)=(-12x)/(x^2+3)^2`.

This question is asking us to find the minimum and maximum values of `f'(x)`.

So we need the derivative of `f'(x)`, the critical numbers for `f'` and then we'll test the critical numbers.

`f''(x) = (36(x^2-1))/(x^2+3)^3`

The critical numbers for `f'` are `-1` and `1`.

On `(-oo,-1)`, `f''` is positive, so `f'` is increasing.
On `(-1,1)`, `f''` is negative, so `f'` is decreasing.
On `(1,oo)`, `f''` is positive, so `f'` is increasing.

`f'` has a local maximum `f'(-1) = 3/4` and a local minimum `f'(1) = -3/4` .

To see that these local extrema are global, note that `f'` has horizontal asymptote `y=0` on both sides.

So on `(-oo,-1)`, `f'` increases from near `0` to a maximum of `3/4` (at `-1`)
`f'` then decreases until it hits a minimum of `-3/4` (at `1`) and then it increases but only to approach its asymptote `y=0`.

Therefore the global maximum for the slope of the tangent to the curve `y = 6/(x^2+3)` is `3/4` and it occurs at `x=-1`. The global minimum for the slope of the tangent to the curve `y = 6/(x^2+3)` is `-3/4` and it occurs at `x=1`.