Question

How do you find the local maximum and minimum values of `f(x) = 2 x + 3 x ^{ -1 }` using both the First and Second Derivative Tests?

Answer 1

Find critical numbers:

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

`f'(x) = 0` at `x = +-sqrt(3/2)` and `f'(x)` DNE at `x=0`.

`0` is not in the domain of `f`, so the only critical numbers ar `+-sqrt(3/2)`

First Derivative Test for Local Extrema

On `(-oo, -sqrt(3/2))`, `f'` is positive
On `(-sqrt(3/2), 0)`, `f'` is negative `" "` so `f(-sqrt(3/2))` is a relative maximum.
On `(0, sqrt(3/2))`, `f'` is negative
On `(sqrt(3/2), oo)`, `f'` is positive `" "` so `f(sqrt(3/2))` is a relative minimum.

Second Derivative Test for Local Extrema

`f''(x) = 6/x^3`

`f''(-sqrt(3/2))` is negative, so `f(-sqrt(3/2))` is a relative maximum.

`f''(-sqrt(3/2))` is positive, so `f(-sqrt(3/2))` is a relative minimum.

Finish by doing arithmetic

Find the values by doing the arithmetic `f(-sqrt(3/2))` and `f(sqrt(3/2))`. (Actually, we only need to do one of these, because `f` is an odd function.)