Given $$X(f)=\frac{2}{(1+j2\pi f)^3}$$ so, $$X^*(f)=\frac{2}{(1-j2\pi f)^3}$$ and a transfer function: $$H(f)=\frac{2(2\pi f)^2 e^{-j2\pi ft}}{(1-j2\pi f)^3}$$ With the following conversion formula: $$\frac{d^n}{dt^n}f(t)=Ft^{-1}[(j2\pi f)^nF(f)]$$
What I got is: $$\frac{d^2}{dt^2}x(-t)$$
Yet the answer is: $$-\frac{d^2}{dt^2}x(-t)$$
I don't understand where does the minus sign come from. I didn't learn fourier transfer properly but the course I am taking asks us to look up the table if we aren't familiar with fourier transform.
Just figured it out. I need to add an extra j in $H(f)$, such that $$H(f)=\frac{2(-1)(j2\pi f)^2 e^{-j2\pi ft}}{(1-j2\pi f)^3}$$ Hence, the extra terms will remains after transformation. $$-\frac{d^2}{dt^2}x(-t)$$