Question

If $f\left(x\right)=x^2+1$ and $g\left(x\right)=2x$, then find the domain of the function
$h\left(x\right)=\frac{(f+g)x}{(f-g)x}$

• A. $\left(-\infty ,\infty \right)-\{-1\}$
• B. $\left(-\infty ,\infty \right)-\left\{1\right\}$
• C. $(-\infty ,-1)\cup (0,\infty )$
• D. $\left(-\infty ,\infty \right)$

Answer 1

The correct option is B $\left(-\infty ,\infty \right)-\left\{1\right\}$

$\frac{(f+g)x}{(f-g)x}=\frac{\left(f\right(x)+g(x\left)\right)}{\left(f\right(x)-g(x\left)\right)}$
$=\frac{x^2+1+2x}{x^2+1-2x}=\frac{{(x+1)}^2}{{(x-1)}^2}$
$\because$ x can take all values except 1 since the denominator cannot be equal to zero.
Therefore, the domain of the given function is $\left(-\infty ,\infty \right)-\left\{1\right\}$