Why Does $f(x) = x\sqrt{x+3}$ Only Have One Critical Point?

Question

I am trying to find the critical points of the function $f(x) = x\sqrt{x+3}$, then by using the First Derivative Test, determine which ones are a local maximum, local minimum, or neither.

Using the product rule, we get $\frac{3(x+2)}{2\sqrt{x+3}}$. So $x=-2$ is a critical point. But isn't $x=-3$ also a critical point because if $x=-3$ then the derivative of $f(x)$ wouldn't exist.

To give an example of why I am thinking this, look at $g(x) = x^{2/3}(6-x)^{1/3}$.

The derivative of $g(x)$ is $\frac{4-x}{x^{1/3}(6-x)^{2/3}}$, where $x$ cannot equal $0$ or $6$.

The critical points for $g(x)$ are $0$, $6$, and $4$.

So why aren't $-2$ and $-3$ the critical points for $f(x)$?

Answer 1

In the hope of clarifying some of the confusion, the critical points of a differentiable function is any value in its domain where the derivative is $0$ or undefined. Given the domain of $f(x)$ is $[-3,\infty)$ and its derivative is:

$$ f'(x) = \dfrac{3(x+2)}{2\sqrt{x+3}} $$

The critical points are:

$-2$ because $-2 \in [-3,\infty)$ and $f'(-2) = 0$.

$-3$ because $-3 \in [-3,\infty)$ and $f'(-3)$ is undefined.

Answer 2

In order for $a$ to be a critical point (at least, under the definition I'm familiar with), $a$ has to be in the interior of the domain of the function (that is, the function has to be defined on either side of $a$); for any $x<-3$, $f(x)$ isn't defined (exercise: why?), so $-3$ can't be a critical point.

Answer 3

you want $\sqrt{-3+3} = 0$ as another point right?

isn't a trying to square root a negative leading to something else entirely like complex numbers.