1. The problem asks why the function $f(x) = (x-1)e^x$ is not periodic. 2. A function $f(x)$ is periodic if there exists a positive number $T$ such that $f(x+T) = f(x)$ for all $x$. 3. Let's test if $f(x)$ is periodic by checking if $f(x+T) = f(x)$ for some $T > 0$: $$f(x+T) = ((x+T)-1)e^{x+T} = (x+T-1)e^{x}e^{T}$$ 4. For $f(x+T) = f(x)$, we need: $$(x+T-1)e^{x}e^{T} = (x-1)e^{x}$$ 5. Dividing both sides by $e^{x}$: $$(x+T-1)e^{T} = x-1$$ 6. Rearranging: $$(x+T-1)e^{T} - (x-1) = 0$$ 7. This must hold for all $x$, so the coefficient of $x$ must be zero: $$xe^{T} + (T-1)e^{T} - x + 1 = 0$$ $$x(e^{T} - 1) + (T-1)e^{T} + 1 = 0$$ 8. For this to be true for all $x$, the coefficient of $x$ must be zero: $$e^{T} - 1 = 0 \\ e^{T} = 1$$ 9. The exponential function $e^{T} = 1$ only when $T=0$, but $T$ must be positive for periodicity. 10. Therefore, no positive $T$ exists such that $f(x+T) = f(x)$, so $f(x)$ is not periodic. Final answer: $f(x) = (x-1)e^x$ is not periodic because the exponential factor $e^x$ does not repeat values for any positive period $T$, and the linear term $(x-1)$ prevents periodicity.