Q.Select the correct answer from the given alternatives. $f(x) = \dfrac{x^2-7x+10}{x^2+2x-8}$, for $x \in [-6,-3]$. (A) $f$ is discontinuous at $x=2$ (B) $f$ is discontinuous at $x=-4$ (C) $f$ is discontinuous at $x=0$ (D) $f$ is discontinuous at $x=2$ and $x=-4$

$f(x)=\dfrac{x^2-7x+10}{x^2+2x-8}$ for $x\in[-6,-3]$. Factor: $x^2-7x+10=(x-2)(x-5)$, and $x^2+2x-8=(x+4)(x-2)$. So for $x\ne2$, $f(x)=\dfrac{(x-2)(x-5)}{(x+4)(x-2)}=\dfrac{x-5}{x+4}$.

Within the given interval $[-6,-3]$, the value $x=2$ never occurs (it is outside this interval), so that removable factor is irrelevant here. What matters is where the simplified denominator $x+4$ vanishes: at $x=-4$, which is inside $[-6,-3]$. …

Factor both numerator and denominator fully; only a factor that appears in both is removable, so check separately which zero of the denominator survives after cancellation — that surviving zer …

Assuming that because $(x-2)$ cancels somewhere in the expression, every discontinuity of the original formula must be removable — the surviving, non-cancelling factor $( …