Q.Discuss the continuity of the following function at the point(s) or on the interval indicated against it: $f(x) = 2x^2-2x+5$, for $0 \le x \le 2$, $= \dfrac{1-3x-x^2}{1-x}$, for $2 < x < 4$, $= \dfrac{x^2-25}{x-5}$, for $4 \le x \le 7$ and $x \ne 5$, $= 7$ for $x=5$.

$f(x)=2x^2-2x+5$ on $[0,2]$; $=\dfrac{1-3x-x^2}{1-x}$ on $(2,4)$; $=\dfrac{x^2-25}{x-5}$ on $[4,7]$, $x\ne5$; $f(5)=7$.

At $x=2$: $f(2)=8-4+5=9$ (first piece). Right-hand limit: $\dfrac{1-6-4}{1-2}=\dfrac{-9}{-1}=9$. These match, so $f$ is continuous at $x=2$.

At $x=4$: Left-hand limit (second piece): $\dfrac{1-12-16}{1-4}=\dfrac{-27}{-3}=9$. Third piece at $x=4$: $\dfrac{16-25}{4-5}=\dfrac{-9}{-1}=9$; since this piece is itself continuous away from $x=5$, the right-hand limit also equals $9$. So $f$ is continuous at $x=4$ too. …

Check each junction of the three-piece function in turn (here $x=2$ and $x=4$ both turn out fine), and separately simplify the piece containi …

Stopping after verifying the two 'obvious' junctions ($x=2,4$) look fine and failing to also check the special interior p …