Questions: 1) Find the domain and range of a) f(x)=1/(sqrt(9-x^2)); b) g(x)=(2x-3)/(x-1).

Solution Steps

To find the domain, determine the values of \( x \) for which the expression under the square root is positive and the denominator is not zero:

1. The expression under the square root must satisfy \( 9 - x^{2} > 0 \).
2. Solve \( 9 - x^{2} > 0 \): \[ 9 - x^{2} > 0 \implies x^{2} < 9 \implies -3 < x < 3. \] Thus, the domain of \( f(x) \) is \( (-3, 3) \).

To find the range, analyze the behavior of \( f(x) \):

1. As \( x \) approaches \( -3 \) or \( 3 \), \( \sqrt{9 - x^{2}} \) approaches \( 0 \), making \( f(x) \) approach \( +\infty \).
2. At \( x = 0 \), \( f(0) = \frac{1}{\sqrt{9 - 0}} = \frac{1}{3} \).
3. Since \( \sqrt{9 - x^{2}} \) is always positive and decreases as \( |x| \) increases, \( f(x) \) increases as \( |x| \) increases. Thus, the range of \( f(x) \) is \( \left[ \frac{1}{3}, +\infty \right) \).

To find the domain, determine the values of \( x \) for which the denominator is not zero:

1. The denominator \( x - 1 \neq 0 \), so \( x \neq 1 \). Thus, the domain of \( g(x) \) is \( (-\infty, 1) \cup (1, +\infty) \).

Final Answer