Finding distribution of $\frac{1}{\sqrt X}$ when $X$ is uniform on $[1,9]$ I haven't been able to figure this question out and have been googling for about a good hour so thought I should post this here.

Let $X$ be a $\mathrm{uniform}(1,9)$ random variable, i.e. it has the probability density function
  $$
  f_X(x) =
\begin{cases}
1/8,  & \text{if 1 $\le$  $x$ $\le$ 9,} \\
0, & \text{otherwise.}
\end{cases}
$$ 
  Find the Cumulative Distribution function and Probability density function of
  $$Y = \frac {1}{\sqrt{X}}.$$

My wrong attempt went as follows:
First I determined the range of $ Y $ to be $y \in [\frac13,1] $ by filling in the boundrairies 1 and 9 into $ Y $ 
Calculate the CDF by taking the close integral over the interval $ [1,x]$
$$ CDF: \int_1^x \frac18 du =  \frac18(x-1)$$
Then I find CDF of $ Y $ $$  F_y = P(Y \le y )  =  P( \frac{1}{\sqrt{X}} \le y ) = P(X \le \frac{1}{y^2} ) = \frac18(\frac{1}{y^2} -1) $$
But this solutions seems to be wrong since it doesn't add up to 1 when integrated over its range $ [\frac13,1]$
If someone could give me an explanaiting where I made an error that be great!
 A: Mistake is in the line $\mathbb P(\frac{1}{\sqrt{X}} \le y) = \mathbb P(X \le \frac{1}{y^2}),$ because due to $\sqrt{X},y > 0$, we have:
$ \frac{1}{\sqrt{X}} \le y  \iff 1 \le \sqrt{X}y \iff \frac{1}{y} \le \sqrt{X} \iff \frac{1}{y^2} \le X,$ so it should be $P(\frac{1}{\sqrt{X}} \le y) = \mathbb P(X \ge \frac{1}{y^2})$
As you calculated $F_X(t) = \frac{x-1}{8}\chi_{[1,9]}(t) $ and correcting your error, we get $($for $y \in [\frac{1}{3},1])$:
$F_Y(y) = \mathbb P(\frac{1}{\sqrt{X}} \le y) = \mathbb P(X \ge \frac{1}{y^2}) = 1 - \mathbb P(X < \frac{1}{y^2}) = 1 - \mathbb P(X \le \frac{1}{y^2}) = 1 - F_X(\frac{1}{y^2}) = 1 - \frac{1}{8}(\frac{1}{y^2}-1) = \frac{1}{8}(9-\frac{1}{y^2}) $
So we get the CDF of $Y$: $F_Y(t) = \frac{1}{8}(9-\frac{1}{t^2})\chi_{[\frac{1}{3},1]}(t)$
Now PDF, as we know since CDF is continuosly differentiable (piecewise at least) it holds, that $f_Y(y) = F_Y'(y)$ for every point in which $F_Y'$ exists (we can put whatever we want in the rest, since $PDF$ is up to the set of measure zero).
Differentiating gives us PDF : $f_Y(y) = \frac{1}{4y^3} \chi_{(\frac{1}{3},1)}(y)$ (Note that you could have included points $\frac{1}{3}$ and/or $1$ in characteristic function.
