This is a continuous time and signal question:
Suppose $h(t)$ is a causal signal and has the even part $he(t)$ given by
$he(t) = t[u(t)−u(t −1)]+u(t −1)$ for $t > 0$.
Find $h(t)$ for all $t$.
I know that $h(t)=he(t) + ho(t)$, and I think you can use a graphing utility to obtain $ho(t)$ (I think), and once you obtain $ho(t)$ you can get $h(t)$ but my problem is I'm bad at unit pulses so I don't understand the $u(t)$ parts. I really need an explanation of how to solve this problem.
$h_e(t)$ is given for $t>0$. You should find the it for $t<0$.
Since it is even, $$h_e(-t)=h_e(t)\Rightarrow h_e(t)=-t[u(-t)−u(-t −1)]+u(-t −1), t<0$$ This is what your $h_e(t)$ looks like:
A causal signal is a signal which is zero when $t<0$. This is what your $h_o(t)$ should do.
For $t<0$ set
$$h_o(t)=-h_e(t) =t[u(-t)−u(-t −1)]-u(-t −1), t<0$$ But since $h_o(t)$ is odd, for $t>0$ change the sign. So it is indeed the same $h_e(t)$ for $t>0$: $$h_o(t)=t[u(t)−u(t −1)]+u(t −1),t>0$$
Your $h_o(t)$ looks like this
and the overall signal $h(t)$ is sum of the four parts above which is
All in all, it is easy to check that $$ \bbox[0.5em,#efe,border:0.1em groove navy]{\ h(t)=2t[u(t)−u(t −1)]+2u(t −1)} $$
You can verify it is a causal signal (it is zero for $t<0$)