I have a second derivative, which is $$\frac{(11x-24)}{4\sqrt{(x-3)^5}\sqrt{(x-2)^3}}$$
Now I am supposed to find the concativity. But how am I supposed to find it, if this function's domain is $(3;+\infty)$? Plugging in, for example, $4$, gives me a positive value so I know that function concaves upwards.
I also tried factoring out $(x-3)^2(x-2)$, which leaves me with $\sqrt{(x-2)(x-3)}$ in the denominator, changing the domain to $(-\infty;2)\bigcup(3;+\infty)$
But then there is another problem. In the interval of $(-\infty;2)$ the function concaves down, but plugging in, for example, $0$, gives me that 2nd derivative is actually positive, which doesn't make sense.
So my questions are: How do i deal with the domain of 2nd derivative (specifically in this example) and the other problem that is described right above?
The starting function was $x\sqrt\frac{x-2}{x-3}$
Let $f(x)=x\sqrt{\frac{x-2}{x-3}}.$
Thus, the domain is $(-\infty,2]\cup(3,\infty)$.
$$f''(x)=\frac{11x-24}{4(x-3)^2\sqrt{(x-2)^3(x-3)}}.$$ We see that $f''(x)<0$ for all $x<2$ and $f''(x)>0$ for all $x>3$, which gives
$f$ is a concave function on $(-\infty,2)$.