Answer to Question #97138 in Calculus for Anand

Question

Find the maximum possible domain and corresponding range of the function f defined by f(x) = {√(1–x²)/(x–2)}

Answer 1

ANSWER: Domain of f is the set   $(-\infty,-1]\bigcup [1,2)$ ; range of f is the set $[0,+\infty)$

EXPLANATION. If $x>2$ , then the denominator of the fraction is positive and the numerator is negative. Therefore the fraction $\frac{1-x^{2}}{x-2}<0$ . If $x<2$ , then the domain of the function f consists of x, for which $1-{ x }^{ 2 }\le 0$ , hence $x\in\left( - \infty,-1 \right ] \bigcup\left[1,+\infty \right )$ . Thus, the domain of the function is the set $\left \{ (-\infty,-1]\bigcup [1,+\infty) \right \} \bigcap (-\infty,2) =$ $(-\infty,-1]\bigcup [1,2)$

The range of the function is the set $[0,+\infty)$ , since for any $a\geqslant0$ the equation f(x)=a has a solution ,that belongs to [1,2). This follows from equivalent equalities :$x\in [1,2)\quad ,\sqrt { \frac { 1-{ x }^{ 2 } }{ x-2 } } =a\quad \Leftrightarrow \quad \frac { 1-{ x }^{ 2 } }{ x-2 } ={ a }^{ 2 }\quad \Leftrightarrow { \quad x }^{ 2 }+{ a }^{ 2 }x-2{ a }^{ 2 }-1=0\$ $\Rightarrow x=\frac { -{ a }^{ 2 } }{ 2 } +\sqrt { \frac { { a }^{ 4 } }{ 4 } +2{ a }^{ 2 }+1 } \quad$