Question

The domain of the function $f\left(x\right)=\frac{1+2\left(x+4\right)^{-0.5}}{2-\left(x+4\right)^{0.5}}+5\left(x+4\right)^{0.5}$ is

• A. $R$
• B. $(-4, 4)$
• C. $\left(0, \infty\right)$
• D. $\left(-4, 0\right) \cup\left(0, \infty\right)$

Answer 1

Answer: (D)

$\left(-4, 0\right) \cup\left(0, \infty\right)$

Answer 2

$f\left(x\right)=\frac{1+2\left(x+4\right)^{-0.5}}{2-\left(x+4\right)^{0.5}}+5\left(x+4\right)^{0.5}$ $=\frac{1+2\cdot \frac{1}{\sqrt{x+4}}}{2-\sqrt{x+4}}+5 \sqrt{x+4}=\frac{\sqrt{x+4}+2}{\sqrt{x+4}\left(2-\sqrt{x+4}\right)}+5\sqrt{x+4}$ $f\left(x\right)$ is defined if $x + 4 > 0$ and $2-\sqrt{x+4}\ne 0$. $\therefore f\left(x\right)$ is defined if $x >\- 4$ and $x \ne 0$ $\therefore D\left(f\right)=\left(-4,0\right) \cup\left(0, \infty\right)$.