Solution:** To find the range of \(f(x) = \frac{2x^2 - 3x + 1}{x^2 + 1}\), set \(y = f(x)\) and solve for \(x\):

["Finding the Range of ( f(x) = \frac{2x^2 - 3x + 1}{x^2 + 1} ): A Step-by-Step Algebraic Solution", "Understanding the range of a rational function is essential in calculus, graphing, and applied mathematics. In this article, we explore how to systematically find the range of the function:\n[\nf(x) = \frac{2x^2 - 3x + 1}{x^2 + 1}\n]\nby setting ( y = f(x) ) and solving for ( x ) in terms of ( y ).", "---", "### Step 1: Set ( y = f(x) )", "Begin by defining ( y ) equal to the function:\n[\ny = \frac{2x^2 - 3x + 1}{x^2 + 1}\n]", "Our goal is to determine all real values of ( y ) for which this equation has real solutions for ( x ).", "---", "### Step 2: Eliminate the denominator", "Multiply both sides by ( x^2 + 1 ) (which is always positive, so no division by zero):\n[\ny(x^2 + 1) = 2x^2 - 3x + 1\n]", "Expanding the left-hand side:\n[\nyx^2 + y = 2x^2 - 3x + 1\n]", "---", "### Step 3: Rearrange into a quadratic in ( x )", "Bring all terms to one side:\n[\nyx^2 - 2x^2 + 3x + y - 1 = 0\n]", "Factor out ( x^2 ):\n[\n(y - 2)x^2 + 3x + (y - 1) = 0\n]", "This is a quadratic equation in the form:\n[\nAx^2 + Bx + C = 0\n]\nwhere\n[\nA = y - 2, \quad B = 3, \quad C = y - 1\n]", "---", "### Step 4: Apply the condition for real solutions", "For the quadratic to have real solutions in ( x ), the discriminant must be non-negative:\n[\n\Delta = B^2 - 4AC \geq 0\n]", "Compute the discriminant:\n[\n\Delta = 3^2 - 4(y - 2)(y - 1) = 9 - 4(y - 2)(y - 1)\n]", "Expand the product:\n[\n(y - 2)(y - 1) = y^2 - 3y + 2\n]", "So,\n[\n\Delta = 9 - 4(y^2 - 3y + 2) = 9 - 4y^2 + 12y - 8 = -4y^2 + 12y + 1\n]", "---", "### Step 5: Solve the inequality ( \Delta \geq 0 )", "We solve:\n[\n-4y^2 + 12y + 1 \geq 0\n]", "Multiply both sides by (-1) (reversing the inequality):\n[\n4y^2 - 12y - 1 \leq 0\n]", "Solve the quadratic equation ( 4y^2 - 12y - 1 = 0 ) using the quadratic formula:\n[\ny = \frac{12 \pm \sqrt{(-12)^2 - 4(4)(-1)}}{2(4)} = \frac{12 \pm \sqrt{144 + 16}}{8} = \frac{12 \pm \sqrt{160}}{8}\n]", "Simplify ( \sqrt{160} = \sqrt{16 \cdot 10} = 4\sqrt{10} ):\n[\ny = \frac{12 \pm 4\sqrt{10}}{8} = \frac{3 \pm \sqrt{10}}{2}\n]", "So, the roots are:\n[\ny_1 = \frac{3 - \sqrt{10}}{2}, \quad y_2 = \frac{3 + \sqrt{10}}{2}\n]", "Since the quadratic ( 4y^2 - 12y - 1 ) opens upwards, the inequality ( 4y^2 - 12y - 1 \leq 0 ) holds between the roots:\n[\n\frac{3 - \sqrt{10}}{2} \leq y \leq \frac{3 + \sqrt{10}}{2}\n]", "---", "### Step 6: Conclude the range", "Thus, the range of ( f(x) ) is:\n[\n\left[ \frac{3 - \sqrt{10}}{2}, \frac{3 + \sqrt{10}}{2} \right]\n]", "This realistic interval captures all possible output values of ( f(x) ) as ( x ) varies over all real numbers, derived directly from the condition that the equation in ( x ) has real solutions.", "---", "### Why This Method Works", "By expressing ( y ) in terms of ( x ) and ensuring the resulting quadratic in ( x ) has real solutions, we systematically constrain ( y ) to values where the rational function is defined and attainable. This method avoids graphical guesswork and provides a rigorous analytical solution.", "---", "### Final Thoughts", "Finding the range of rational functions often requires algebraic manipulation of the equation, emphasizing the discriminant’s role. Mastering this technique enhances problem-solving in advanced algebra, optimization, and engineering applications.", "---", "Keywords:\n( f(x) = \frac{2x^2 - 3x + 1}{x^2 + 1} ), range of rational function, algebra, discriminant method, solving rational equations, real solutions, calculus preparation.", "Meta Description:\nLearn how to find the range of ( f(x) = \frac{2x^2 - 3x + 1}{x^2 + 1} ) by setting ( y = f(x) ), expressing as a quadratic in ( x ), and analyzing the discriminant. Step-by-step solution with algebraic explanation.", "---", "References & Further Reading:\n- Algebraic methods in solving rational equations\n- Discriminant condition for real roots\n- Graphing rational functions and their domains and ranges"]