Question

The range of $f\left(x\right)=\frac{x^2-9}{x-3}$ is

• A. $(-\infty ,6)\cup (6,\infty )$
• B. $(-\infty ,\infty )$
• C. $R-\left\{9\right\}$
• D. $(0,\infty )$

Answer 1

Answer: (A)

$(-\infty ,6)\cup (6,\infty )$

$f\left(x\right)=\frac{x^2-9}{x-3}$
$x-3\ne 0$
$x\ne 3$
$f\left(x\right)=x+3,x\ne 3$
Range of $f\left(x\right)$ is $(-\infty ,\infty )-6$