Discuss the continuity of the following function at the point or … - Vidyadip

Question

Discuss the continuity of the following function at the point or on the interval indicated against them. If the function is discontinuous, identify the type of discontinuity and state whether the discontinuity is removable. If it has a removable discontinuity, redefine the function so that it becomes continuous:
$f(x)=\frac{(x+3)\left(x^2-6 x+8\right)}{x^2-x-12}$

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Answer

$\begin{aligned}
& f(x)=\frac{(x+3)\left(x^2-6 x+8\right)}{x^2-x-12} \\
& f(x)=\frac{(x+3)\left(x^2-6 x+8\right)}{(x-4)(x+3)}
\end{aligned}$
$\therefore \mathrm{f}(\mathrm{x})$ is not defined at for $\mathrm{x}=4$ and $\mathrm{x}=-3$
$\therefore$ The domain of function $f=R-\{-3,4\}$ for $x \neq-3,4$
$\begin{aligned}
& f(x)=\frac{(x+3)(x-2)(x-4)}{(x-4)(x+3)} \\
& \therefore f(x)=x-2, x \neq-3,4 \\
& \therefore f(-3)=-5 \text { and } f(4)=2
\end{aligned}$
$f(x)$ is discontinuous $x=4$ and $x=-3$
This discontinuity is removable.
$\therefore \mathrm{f}(\mathrm{x})$ can be redefined as
$f(x)=\frac{(x+3)\left(x^2-6 x+8\right)}{x^2-x-12}$
$=-5$, for $x=-3$

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