Let $X\approx \mathcal N(\mu, \sigma ^2)$ be a Gaussian random variable. Let $a,b \in \Bbb R$, show that $aX+b$ is also a Gaussian random variable.
I have the following formula, for a continuous bijection $\phi:\Bbb R \to \Bbb R$, we have that $$ f_{\phi(X)}(x)=\frac{1}{\lvert \phi'(\phi^{-1}(x))\rvert} f_X(\phi^{-1}(x))$$
With $\phi(x)=ax+b$, $\phi^{-1}(x)=\frac{x-b}a$, I have that $$f_{aX+b}(x)=\frac1{\lvert x-b \rvert}\frac{1}{\sqrt{2 \pi \sigma^2}}\exp \bigg(-\frac{(x-(b+a\mu))^2}{2(a\sigma)^2}\bigg)$$
The $\frac1{\lvert x-b \rvert}$ is really bothering me and seems to be unremovable, how can I finish this computation ?
Notice, that $\phi'(y) = a$ for any $y$ in particular $$\frac{1}{\phi'(\phi^{-1}(x))} = \frac{1}{|a|} = \frac{1}{\sqrt{a^2}}.$$ Can you finish the computation from here?