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