Find the range of real-valued rational functions using different techniques such as inverse functions and quadratic discriminants.
Examples with Solutions
Example 1
Find the range of function \( f \) defined by:
\[ f(x) = \frac{x + 1}{2x - 2} \]View Solution
Write the given function as an equation:
\[ y = \frac{x + 1}{2x - 2} \]Solve the equation for \( x \):
\[ y(2x - 2) = x + 1 \] \[ 2yx - 2y = x + 1 \] \[ 2yx - x = 2y + 1 \] \[ x(2y - 1) = 2y + 1 \] \[ x = \frac{2y + 1}{2y - 1} \]The above expression for \( x \) in terms of \( y \) shows that \( x \) is real for all real values of \( y \) except \( \frac{1}{2} \), since \( y = \frac{1}{2} \) makes the denominator \( 2y - 1 = 0 \).
Hence, the range of \( f \) is given by the interval:
\[ (-\infty, \frac{1}{2}) \cup (\frac{1}{2}, +\infty) \]
Example 2
Find the range of function \( f \) defined by:
\[ f(x) = \frac{x + 2}{x^2 - 9} \]View Solution
Write the given function as an equation:
\[ y = \frac{x + 2}{x^2 - 9} \]Rewrite the equation as a quadratic equation in standard form for \( x \):
\[ y(x^2 - 9) = x + 2 \implies yx^2 - x - 9y - 2 = 0 \]Find the discriminant of the quadratic equation:
\[ \Delta = (-1)^2 - 4(y)(-9y - 2) = 36y^2 + 8y + 1 \]Using the quadratic formula, the solutions for \( x \) are:
\[ x_{1,2} = \frac{1 \pm \sqrt{36y^2 + 8y + 1}}{2y} \]The solutions are real if \( 36y^2 + 8y + 1 \geq 0 \) and \( y \neq 0 \). The discriminant of \( 36y^2 + 8y + 1 \) is \( 8^2 - 4(36)(1) = -80 \), which is negative. Therefore, \( 36y^2 + 8y + 1 \) is always positive for all real numbers.
For \( y = 0 \), substituting \( y = 0 \) into \( \frac{x + 2}{x^2 - 9} = 0 \) gives \( x = -2 \), meaning \( y = 0 \) is also in the range. Thus, the range of \( f \) is:
\[ (-\infty, +\infty) \]
Example 3
Find the range of function \( f \) defined by:
\[ f(x) = \frac{x + 2}{x^2 + 1} \]View Solution
Write the function as an equation:
\[ y = \frac{x + 2}{x^2 + 1} \]Rewrite as a quadratic equation in \( x \):
\[ y(x^2 + 1) = x + 2 \implies yx^2 - x + y - 2 = 0 \]Solve for \( x \) using the quadratic formula:
\[ x_{1,2} = \frac{1 \pm \sqrt{1 - 4y(y - 2)}}{2y} = \frac{1 \pm \sqrt{1 - 4y^2 + 8y}}{2y} \]The solutions are real if the radicand is non-negative and \( y \neq 0 \). Solving \( 1 - 4y^2 + 8y \geq 0 \) gives the interval:
\[ y = \frac{8 \pm \sqrt{64 + 16}}{8} = \frac{8 \pm \sqrt{80}}{8} \]Simplify and write the range as:
\[ \left[ 1 - \frac{\sqrt{5}}{2}, 1 + \frac{\sqrt{5}}{2} \right]\]
Example 4
Find the range of function \( f \) defined by:
\[ f(x) = \frac{x^2 + 2}{x^2 + 1} \]View Solution
Write the function as an equation:
\[ y = \frac{x^2 + 2}{x^2 + 1} \]Rewrite as a quadratic equation in \( x \):
\[ y(x^2 + 1) = x^2 + 2 \implies x^2(y - 1) = 2 - y \]Solve for \( x \):
\[ x_{1,2} = \pm\sqrt{\frac{2 - y}{y - 1}} \]The solutions are real if the radicand is non-negative:
\[ \frac{2 - y}{y - 1} \geq 0 \]Solving this inequality gives:
\[ 1 < y \leq 2 \]Thus, the range of the function is given by the interval:
\[ (1, 2] \]
Example 5
Find the range of function \( f \) defined by:
\[ f(x) = \frac{1}{x^2 - 1} \]View Solution
Write the function as an equation:
\[ y = \frac{1}{x^2 - 1} \]Rewrite as a quadratic equation in \( x \):
\[ y(x^2 - 1) = 1 \implies x^2 y = y + 1 \implies x_{1,2} = \pm\sqrt{\frac{y + 1}{y}} \]For solutions to be real, the radicand must be non-negative:
\[ \frac{y + 1}{y} \geq 0 \]The solution set to this inequality is:
\[ (-\infty, -1] \cup (0, +\infty) \]
Example 6
Find the range of function \( f \) defined by:
\[ f(x) = \frac{2x^2 - 1}{x + 1} \]View Solution
Write the function as an equation:
\[ y = \frac{2x^2 - 1}{x + 1} \]Rewrite as a quadratic equation in \( x \):
\[ y(x + 1) = 2x^2 - 1 \implies 2x^2 - yx - y - 1 = 0 \]Solve for \( x \) using the quadratic formula:
\[ x_{1,2} = \frac{y \pm \sqrt{y^2 + 8y + 8}}{2} \]The solutions are real if the radicand is non-negative:
\[ y^2 + 8y + 8 \geq 0 \]Solving this inequality gives the range:
\[ (-\infty, -4 - 2\sqrt{2}] \cup [-4 + 2\sqrt{2}, +\infty) \]