Question

What are the values and types of the critical points, if any, of `f(x)=x / (x^2 + 4)`?

Answer 1

`x=+-2`-Stationary

Explanation:

Let `f=x and g=x^2+4`

`f'=1 and g'=2x`

`f'(x)=(gf'-fg')/g^2`

`f'(x)=(x^2+4-2x^2)/(x^2+4)^2 =0`

`(4-x^2)/(x^2+4)^2 =0`

Denominator goes to zero when we multiply to the other side to solve for x.

`4-x^2=0`

`4=x^2`

`x=+-2`

Since the denominator will never equal to zero we only have one type of critical points and that is stationary

Answer 2

The critical points are `-2` and `2`. Furthermore, `-2` is the location of a local minimum and `2` is the location of a local maximum.

Explanation:

There are several different uses of "critical point" and "types of critical point".

Meaning of "critical point"
In the terminology I was taught (and that I teach), a critical point of a function `f`, is a point (call it `c`) in the domain of the function with either `f'(c)=0` or f'(c) does not exist.

With that definition, we proceed:

`f(x) = x/(x^2+4)` has domain `(-oo,oo)`

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

`f'` is never undefined and `f'(x) = 0` at `x=+- 2` (both of which are in the domain of `f`.

"types" of critical points

Again, using the terminology I am familiar with, a critical point may be the location of a local minimum, a local maximum or neither. (In addition to "local" I also use "relative" -- they mean the same thing.)

Applying the first derivative text we find that there is a local minimum at `-2` (If you prefer, we can say "at `x=-2`")
That minimum is `f(-2) = -1/4`

Applying the first derivative text we find that there is a local maximum at `2`.
That minimum is `f(2) = 1/4`.

Other uses of the words

I have seen some who define a critical point as a point on the graph of a function where the derivative is `0` or fails to exist.

The would say that the critical points are `(-2, -1/4)` and `(2, 1/4)`,

Some would also say that `(-2, -1/4)` is a minimum point and `(2, 1/4)` is a maximum point.