Fourier transforming something of the form x(at+b) when the transform of x(t) is known is slightly more difficult than one would think. For instance, if we want to transform x(-t-1), should we shift or invert first? Not being sure, perhaps it would be better to use the definition of a Fourier transform to solve this.
$ F(x(t))=\int_{-\infty}^\infty x(t)e^{-j\omega t} dt = \chi(\omega) $
So, $ F(x(at+b))=\int_{-\infty}^\infty x(at+b)e^{-j\omega t} dt $
Let $ \tau = at+b. $ This implies $ t=\frac{\tau-b}{a} $ and $ dt = \frac{d\tau}{a} $.
Case 1: Consider a > 0.
$ t = -\infty \Rightarrow \tau = -\infty $; $ t = \infty \Rightarrow \tau = \infty $
$ F(x(at+b))=\int_{-\infty}^\infty x(\tau)e^{-j\omega \frac{\tau-b}{a}} \frac{d\tau}{a} $
$ =\frac{1}{a} e^{j\omega \frac{b}{a}} \int_{-\infty}^\infty x(\tau)e^{-j\frac{\omega}{a}\tau} d\tau $
$ =\frac{1}{a} e^{j\omega \frac{b}{a}} \chi(\frac{\omega}{a}) $
Case 2: Consider a < 0.
$ t = -\infty \Rightarrow \tau = \infty $; $ t = \infty \Rightarrow \tau = -\infty $
$ F(x(at+b))=\int_{\infty}^{-\infty} x(\tau)e^{-j\omega \frac{\tau-b}{a}} \frac{d\tau}{a} $
$ =\frac{-1}{a} e^{j\omega \frac{b}{a}} \int_{-\infty}^\infty x(\tau)e^{-j\frac{\omega}{a}\tau} d\tau $
$ =\frac{1}{(-a)} e^{j\omega \frac{b}{a}} \chi(\frac{\omega}{a}) $
So, combining the two cases:
$ F(x(at+b))=\frac{1}{|a|} e^{j\omega \frac{b}{a}} \chi(\frac{\omega}{a}) $
