Finding probability distribution function when given a different continuous random variable. Question:

The continuous random variable $T$ is distributed uniformly on the interval $[2, 4].$
Given that $V = \sqrt{2T - 7}$  find the probability distribution function for $V.$


So far I've been able to do the following:

$P(T = t) = 0.5$, $P(T \leq t) = 0.5t - 1$
Rearrange and plug in from $P(T<=t)$:
$P(V<=v)= P(\sqrt{2T-7} \leq v) = P(T\leq (v^2+7)/4) = 0.5(\frac{v^2+7}2) - 1 $
$ = \frac{1}4v^2+\frac{3}4$
Given that the limits of t are between 2 and 4, from my understanding, the limits of v should then be $ 0\leq v\leq 1$, since $ \sqrt{4 \cdot 2 - 7} = 1$.
However, this means than if I plug in $v = 0$ into my equation $P(V \leq v) = \frac{1}4v^2+\frac{3}4$, I get $P(V\leq v) = \frac{3}4 $, which is wrong.
However, I'm not sure where I have gone wrong in my working,



Any help would be appreciated,
Many thanks

$V$ is not "well-defined." It is neither continuous nor discrete. If $3.5<T<4$, then $f(v)=2v, 0<v<1$. This integrates to 1. But it doesn't make sense to say that 1/4 of the time, V has this pdf, and the other 3/4 of the time, V is undefined, because it is T that lies in 3.5<T<4 1/4 the time, but this doesn't mean that the pdf of V will be that 1/4 the time; in fact, it is always that when V has certain values. Or if you made it proportional to the actual probability, then $f(v)=\frac v 2, 0<v<1$, but what would you do with values outside this range. They would have to be 0. But then this isn't a pdf, and also doesn't tell you when V is undefined.
