Question

Find the domain of the function \[f\left( x \right) = {\log _{10}}\left( {\sqrt {x - 4} + \sqrt {6 - x} } \right)\].
A. \[\left[ {4,6} \right]\]
B. \[\left( { - \infty ,6} \right)\]
C. \[\left[ {2,3} \right]\]
D. None of these

Answer 1

Hint We know that the value of an expression under the square root function is never negative. If the value of expression under the square root function is negative, then it becomes a complex number. There are two radical functions. So they must be greater than or equal to zero. From this, we can make two inequalities. The intersection of the solution of the inequalities in the domain of the function.

Complete step by step solution
Given function is \[f\left( x \right) = {\log _{10}}\left( {\sqrt {x - 4} + \sqrt {6 - x} } \right)\].
There are two radicals in the given function.
The radicals are \[\sqrt {x - 4} \] and \[\sqrt {6 - x} \].
Since the value of an expression under square root function is never negative.
So, \[x - 4\] and \[6 - x\] must be greater than or equal to zero.
Therefore
\[x - 4 \ge 0\] and \[6 - x \ge 0\]
Now we will solve the above inequalities
\[x \ge 4\] and \[6 \ge x\]
Now we will apply the interval notation.
\[x \in \left[ {4,\infty } \right)\] and \[x \in \left( { - \infty ,6} \right]\]
So, the intersection of \[x \in \left[ {4,\infty } \right)\] and \[x \in \left( { - \infty ,6} \right]\] is \[\left[ {4,6} \right]\].
Hence the correct option is option A.

Note: Students often make a common mistake to solve the inequality \[6 - x \le 0\]. They subtract both sides by \[ - 6\]. They solve it like \[6 - x - 6 \le 0 - 6\] \[ \Rightarrow - x \le - 6\] \[ \Rightarrow x \le 6\]. The solution of the inequality is \[\left[ {-\infty,6 } \right)\]. This is the wrong solution.