Question

Find the domain of the function $y=\sqrt{16x-{{x}^{5}}}+{{\log }_{\dfrac{1}{2}}}\left( {{x}^{2}}-1 \right)$.

Answer 1

Hint: First we will consider the given equation as two parts one is $\sqrt{16x-{{x}^{5}}}$ and the second one is ${{\log }_{\dfrac{1}{2}}}\left( {{x}^{2}}-1 \right)$. To find the domain of the given function we need to calculate the interval where the two parts of the given function are defined. We know that the function which is square root is defined when that function is given values greater than or equal zero. From this we will calculate the interval where the part $\sqrt{16x-{{x}^{5}}}$ is defined. We also know that the logarithmic function is defined when the function value is greater than zero. From this we will calculate the interval where the part ${{\log }_{\dfrac{1}{2}}}\left( {{x}^{2}}-1 \right)$ is defined. From the obtained two intervals we will calculate the domain of the given function.

Complete step by step answer:
Given that, $y=\sqrt{16x-{{x}^{5}}}+{{\log }_{\dfrac{1}{2}}}\left( {{x}^{2}}-1 \right)$
Consider the part $\sqrt{16x-{{x}^{5}}}$. Now the function $\sqrt{16x-{{x}^{5}}}$ is defined when
$\begin{align}
  & 16x-{{x}^{5}}\ge 0 \\
 & \Rightarrow x\left( 16-{{x}^{4}} \right)\ge 0 \\
 & \Rightarrow 16-{{x}^{4}}\ge 0 \\
 & \Rightarrow {{x}^{4}}\le 16 \\
 & \Rightarrow {{x}^{4}}\le {{\left( \pm 2 \right)}^{4}} \\
 & \Rightarrow x\le \pm 2 \\
\end{align}$
From this the interval where the function $\sqrt{16x-{{x}^{5}}}$ is defined is given by $x\in \left[ -\infty ,-2 \right]\cup \left[ 0,2 \right]...\left( \text{i} \right)$
Now consider the part ${{\log }_{\dfrac{1}{2}}}\left( {{x}^{2}}-1 \right)$. The function ${{\log }_{\dfrac{1}{2}}}\left( {{x}^{2}}-1 \right)$ is defined when
$\begin{align}
  & {{x}^{2}}-1>0 \\
 & \Rightarrow {{x}^{2}}-{{1}^{2}}>0 \\
 & \Rightarrow \left( x+1 \right)\left( x-1 \right)>0 \\
 & \Rightarrow x+1>0\text{ and }x-1>0 \\
 & \Rightarrow x>-1\text{ and }x>1 \\
\end{align}$
From this we can write that the function ${{\log }_{\dfrac{1}{2}}}\left( {{x}^{2}}-1 \right)$ is defined when $x\in \left( -\infty ,-1 \right)\cup \left( 1,\infty \right)...\left( \text{ii} \right)$
If $y=\sqrt{16x-{{x}^{5}}}+{{\log }_{\dfrac{1}{2}}}\left( {{x}^{2}}-1 \right)$, from equations $\left( \text{i} \right)$ and $\left( \text{ii} \right)$ the domain of $y$ is given by $\left[ -\infty ,-2 \right]\cup \left[ 1,2 \right]$.

Note: While finding the domains and ranges we need to be very careful even the brackets also change our answer. The domain and the ranges are completely depending on the type of the functions we have like functions having square roots have a different domain and the function having the logarithmic has a different domain.