Question

How do you find all local maximum and minimum points using the second derivative test given `y=x+1/x`?

Answer 1

Please see the explanation section below

Explanation:

`y = f(x) = x+1/x`.

Find the critical numbers for `f`

The domain of `f` is `(-oo,0) uu (0,oo)`

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

`y'` fails to exist at `0` which is not in the domain, so is not a critical number.

`y'=0` at `-1` and at `1`. Both are in the domain, so both are critical numbers.

Find the second derivative

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

Apply the test

At `x=-1` we have `y'' = f''(-1) = 2/(-1)^3 < 0`.

Since the second derivative is negative at the critical number `-1`, we conclude that `f(-1)` is a relative maximum.

There is a relative maximum of `-2` at `x=-1`.

I assume that "ralative maximum point" means "point on the graph where `y` is a relative maximum". If so, the answer should be

Relative maximum point: `(-1,-2)`

At `x=1` we have `y'' = f''(1) = 2/(1)^3 > 0`.

Therefore `f(1) = 2` is a relative minimum.

Again, I assume the requested form for the answer is

Relative minimum point: `(1.2)`

Additional note

We have finished answering the question, but the answer may look strange to students. (The relative minimum is greater than the relative maximum.)

Here is the graph of `y = x+1/x`

graph{x+1/x [-10.38, 12.12, -7.335, 3.915]}

Note the discontinuity at `0` that results in two separate branches of the graph. One branch has a maximum of `-2` at `-1` and the other has a minimum of `2` at `1`.