Question #97138
Find the maximum possible domain and corresponding range of the function f defined by f(x) = {√(1–x²)/(x–2)}
1
Expert's answer
2019-10-22T12:23:09-0400


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

EXPLANATION. If x>2x>2 , then the denominator of the fraction is positive and the numerator is negative. Therefore the fraction 1x2x2<0\frac{1-x^{2}}{x-2}<0 . If x<2x<2 , then the domain of the function f consists of x, for which 1x201-{ x }^{ 2 }\le 0 , hence x(,1][1,+)x\in\left( - \infty,-1 \right ] \bigcup\left[1,+\infty \right ) . Thus, the domain of the function is the set {(,1][1,+)}(,2)=\left \{ (-\infty,-1]\bigcup [1,+\infty) \right \} \bigcap (-\infty,2) = (,1][1,2)(-\infty,-1]\bigcup [1,2)

The range of the function is the set [0,+)[0,+\infty) , since for any a0a\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\ x=a22+a44+2a2+1\Rightarrow x=\frac { -{ a }^{ 2 } }{ 2 } +\sqrt { \frac { { a }^{ 4 } }{ 4 } +2{ a }^{ 2 }+1 } \quad


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