Q.Activity: Suppose $f(x) = px + 3$ for $a \le x \le b$, $= 5x^2 - q$ for $b < x \le c$. Find the condition on $p, q$, so that $f(x)$ is continuous on $[a,c]$, by filling in the boxes: $f(b) = \Box$; $\displaystyle\lim_{x\to b^+} f(x) = \Box$; $\therefore pb+3 = \Box - q$; $\therefore p = \dfrac{\Box}{b}$ is the required condition.

Here $f(x)=px+3$ for $a\le x\le b$, and $f(x)=5x^2-q$ for $b<x\le c$. Since $x=b$ belongs to the first piece, $f(b)=pb+3$ — this fills the first box.

The right-hand limit, using the second piece (valid just above $b$), is $\displaystyle\lim_{x\to b^+} f(x)=5b^2-q$ — this fills the second box.

For $f(x)$ to be continuous on $[a,c]$, it must in particular be continuous at $x=b$, i.e. $f(b)=\displaystyle\lim_{x\to b^+} f(x)$:

This fills the third box (with $5b^2-q$ on the right of "$pb+3=$"). …

Fill each box in turn: $f(b)$ from the first piece, the right-hand limit from the second piece, set t …

Forgetting that continuity on the whole closed interval $[a,c]$ only actually needs checking at the single interior junction $x=b$ — the two pieces are individually …