Is the Inverse Calculation for a Piecewise CDF Correct?

[Mogarrr]

Homework Statement

Show that the given function is a cdf (cumulative distribution function) and find $F_X^{-1}(y)$
(c) $F_X(x) = \frac {e^{x}}4$, if $x<0$, and $1-(\frac {e^{-x}}4)$, if $x \geq 0$

Homework Equations

for a strictly increasing cdf, $F_X^{-1}(y) = x \iff F_X(x) = y$

and for a non-decreasing (a.k.a. difficult problem) cdf, $F_X^{-1}(y) = inf \{ x: F_X(x) \geq y \}$

The Attempt at a Solution

It's not so hard to show that F is a cdf. The $lim_{x \to -\infty} F_X(x)= 0$, the $lim_{x \to \infty} F_X(x) = 1$, the function is non-decreasing, and right-continuous.

I have the solution for the inverse, but it doesn't seem right to me. The given solution is

$F_X^{-1}(y) = ln(4y)$ for $0<y< \frac 14$ and $-ln(4(1-y))$ for $\frac 14 \leq y<1$

But this solution doesn't seem to agree with the definition of inverse F or the inverses I found.

so if $y = e^{\frac {x}4}$, then doesn't this imply $x = 4lny$? and doesn't $y= 1 - (e^{\frac {-x}4 })$, imply that $x = -4ln(1-y)$?
For example, using the inverses I have, the set of x's such that $F_X(x) \geq \frac 14$ would include $4ln( \frac 12) \approx -2.772$ and $-4ln(1- \frac 12) \approx 2.772$, and the infimum of the set (greatest number that is less than all other numbers in the set, I think) is -2.772. So I should use the first inverse, $F_X^{-1}(y) = 4lny$, since this function would have the smallest x's.

Am I right here? Please help.

 

[Mogarrr]

so if $y = e^{\frac {x}4}$, then doesn't this imply $x = 4lny$? and doesn't $y= 1 - (e^{\frac {-x}4 })$, imply that $x = -4ln(1-y)$?

I see my mistake now.