Can someone explain how to get this in unit step form?

Expert intro: This question is about writing a piecewise change with the Heaviside unit step function. The key is to match when the expression turns on or changes.

Detailed walkthrough

If the expression changes at $t=3$, then the unit step function should appear as $u(t-3)$.

A standard pattern is:

• before $t=3$, one expression applies
• at and after $t=3$, an additional term is added or subtracted

If you want a function that is $t$ for $t<3$ and changes to $t-1$ for $t\ge 3$, then one correct unit step form is

$t-u(t-3).$

Your guess $t+u(t-3)$ would instead add 1 after $t=3$, so it represents a different change.

Why this works

Since

0, & t<3,\\ 1, & t\ge 3, \end{cases}$$ then $$t-u(t-3)=\begin{cases} t, & t<3,\\ t-1, & t\ge 3. \end{cases}$$ So if the intended function is a downward jump of 1 at $t=3$, the correct form is $$\boxed{t-u(t-3)}.$$ ### 💡 Pitfall guide A common mistake is to use $u(t)$ instead of $u(t-3)$. The shift matters: $u(t-3)$ turns on at $t=3$, while $u(t)$ turns on at $t=0$. Another error is adding a step when the graph actually drops, which requires subtraction. ### 🔄 Real-world variant If the jump size were different, say a drop of $5$, the form would be $t-5u(t-3)$. If the change happened at $t=a$ instead of $t=3$, the step term would become $u(t-a)$. ### 🔍 Related terms unit step function, Heaviside function, piecewise function