Question:
The function $f(x)=\frac{x^{3}+x^{2}-16 x+20}{x-2}$ is not defined for $x=2$. In order to make $f(x)$ continuous at $x=2$, Here $f(2)$ should be defined as
(a) 0
(b) 1
(C) 2
(d) 3
Solution:
Here,
$x^{3}+x^{2}-16 x+20$
$=x^{3}-2 x^{2}+3 x^{2}-6 x-10 x+20$
$=x^{2}(x-2)+3 x(x-2)-10(x-2)$
$=(x-2)\left(x^{2}+3 x-10\right)$
$=(x-2)(x-2)(x+5)$
$=(x-2)^{2}(x+5)$
So, the given function can be rewritten as
$f(x)=\frac{(x-2)^{2}(x+5)}{x-2}$
$\Rightarrow f(x)=(x-2)(x+5)$
If $f(x)$ is continuous at $x=2$, then
$\lim _{x \rightarrow 2} f(x)=f(2)$
$\Rightarrow \lim _{x \rightarrow 2}(x-2)(x+5)=f(2)$
$\Rightarrow f(2)=0$
Hence, in order to make $f(x)$ continuous at $x=2, f(2)$ should be defined as $0 .$