The functions i'm trying to determine the convergence properties of are
a) $f_n(x)=xe^{-nx}$ on $[0,\infty)$
and
b) $f_n(x)=nxe^{-nx}$ on $[0,\infty)$
for (a), $f_n(x)=x/e^{nx}$, it looks to me like this function will converge pointwise to zero as $n \rightarrow \infty$ because $e^{nx}$ will go to infinity and $x$ will be fixed because we are determining the pointwise convergence. I do not know what the intervals of uniform convergence will be, I'd like some help with that actually.
for (b), $f_n(x)=xn/e^{nx}$, I believe this function will also converge pointwise to zero because the exponential decay will decay faster than $n$ will grow. I also need help determining the intervals of uniform convergence for this series.
You are correct that both converge pointwise to zero.
For uniform convergence, let's consider (b). It is not uniformly convergent on $[0,\infty).$ Note that $f_n$ converges uniformly to $f$ on $S$ means $\sup_{x\in S}|f_n(x)-f(x)| \to 0.$ Here our limit is $f(x)=0$ so this means $\sup_{x\in S}|f_n(x)| \to 0.$ We can see that this doesn't hold for $S=[0,\infty)$ since $f_n(x)$ has a maximum of $1/e$ at $x=1/n,$ so $\sup_{x\in [0,\infty)}|f_n(x)| = 1/e$ for all $n$ and this does not converge to zero.
When you consider other intervals $[a,b)$ where $a>0,$ consider that location the maximum of $f_n$ is $1/n$ so will eventually (when $n$ gets large enough) go out of the interval, and the maximum of $f_n$ on $[a,b)$ will occur at $x=a.$ (It may help to sketch the function if this is not obvious.)
Hopefully once you understand (b), (a) will be easier.