Question

Which of the following functions $f$ has/ have a removable discontinuity at the indicated point?

• A. $f\left(x\right)=\frac{x^2-2x-8}{x+2}$ at $x=-2$
• B. $f\left(x\right)=\frac{x-7}{|x-7|}$ at $x=7$
• C. $f\left(x\right)=\frac{x^3+64}{x+4}$ at $x=-4$
• D. $f\left(x\right)=\frac{3-\sqrt{x}}{9-x}$ at $x=9$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $f\left(x\right)=\frac{x^2-2x-8}{x+2}$ at $x=-2$
C $f\left(x\right)=\frac{x^3+64}{x+4}$ at $x=-4$
D $f\left(x\right)=\frac{3-\sqrt{x}}{9-x}$ at $x=9$

$f\left(x\right)=\frac{x^2-2x-8}{x+2}$ at $x=-2$
$=\frac{(x+2)(x-4)}{x+2}$
$=x-4,x\ne 2$

Hence, $f\left(x\right)$ has removable discontinuity at $x=-2$.

Similarly, $f\left(x\right)$ in $\frac{x^3+64}{x+4}$ and $\frac{3-\sqrt{x}}{9-x}$ has removable discontinuity at $x=-4$ and $x=9$ respectively.

$f\left(x\right)=\frac{x-7}{|x-7|}$ at $x=7$
$f\left(x\right)=\{\begin{array}{cc}-1,& x<7\\ 1,& x>7\end{array}$
Hence, $f\left(x\right)$ has non-removable discontinuity at $x=7$.