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

Apr 14, 2016

$x = \pm 2$-Stationary

#### Explanation:

Let $f = x \mathmr{and} g = {x}^{2} + 4$

$f ' = 1 \mathmr{and} g ' = 2 x$

$f ' \left(x\right) = \frac{g f ' - f g '}{g} ^ 2$

$f ' \left(x\right) = \frac{{x}^{2} + 4 - 2 {x}^{2}}{{x}^{2} + 4} ^ 2 = 0$

$\frac{4 - {x}^{2}}{{x}^{2} + 4} ^ 2 = 0$

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

$4 - {x}^{2} = 0$

$4 = {x}^{2}$

$x = \pm 2$

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

Apr 14, 2016

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 $f ' \left(c\right) = 0$ or f'(c) does not exist.

With that definition, we proceed:

$f \left(x\right) = \frac{x}{{x}^{2} + 4}$ has domain $\left(- \infty , \infty\right)$

$f ' \left(x\right) = \frac{\left(1\right) \left({x}^{2} + 4\right) - \left(x\right) \left(2 x\right)}{{x}^{2} + 4} ^ 2 = \frac{4 - {x}^{2}}{{x}^{2} + 4} ^ 2$

$f '$ is never undefined and $f ' \left(x\right) = 0$ at $x = \pm 2$ (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 $f \left(- 2\right) = - \frac{1}{4}$

Applying the first derivative text we find that there is a local maximum at $2$.
That minimum is $f \left(2\right) = \frac{1}{4}$.

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 $\left(- 2 , - \frac{1}{4}\right)$ and $\left(2 , \frac{1}{4}\right)$,

Some would also say that $\left(- 2 , - \frac{1}{4}\right)$ is a minimum point and $\left(2 , \frac{1}{4}\right)$ is a maximum point.