Is $f(x) = x \sqrt{4-x}$ decreasing at $x = 4$?

Question

I am solving a single variable calculus problem and it asks me to determine the decreasing and decreasing intervals of the function $f(x) = x \sqrt{4-x}$. It's pretty clear that from $\left] -\infty , \frac{8}{3} \right[ $ the function is increasing but, since the domain of the function goes only until $x = 4$, should I write that the decreasing interval of the function is $\left] \frac{8}{3} , 4 \right[$ or $\left] \frac{8}{3} , 4 \right]$?

Answer 1

Neither.

The derivative of your function $f$ is $$ f'(x)=\sqrt{4-x}-\frac{x}{2\sqrt{4-x}}=\frac{8-3x}{2\sqrt{4-x}} $$ for $x<4$ (no differentiability at $4$).

The point $x=8/3$ is a local maximum. Hence the function is decreasing over $$ [8/3,4] $$ because, for every $x,y\in [8/3,4]$, if $x<y$ then $f(x)>f(y)$.

The function is increasing over $(-\infty,8/3]$.