91Ó°ÊÓ

Quadratic functions are equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). These functions create a parabola when graphed on a coordinate plane. Understanding their components and how they behave is key to solving inequalities such as the one in the exercise.

In the step-by-step solution, we transformed the inequality \(x + \frac{1}{x} \geq 2\) into a quadratic inequality \(x^2 - 2x + 1 \geq 0\). This expression was then factorized into a perfect square, \((x-1)^2\), which is a special type of quadratic function.

Quadratic functions can be solved by factoring, using the quadratic formula, or completing the square. In this exercise, recognizing \((x - 1)^2\) as a perfect square was crucial to proving that the inequality holds for any positive \(x\). By rearranging and factoring, we demonstrated that the expression on the left side of the inequality is always non-negative, thus confirming the solution.

A reciprocal of a number \(x\) is simply \(\frac{1}{x}\). If you multiply a number by its reciprocal, you get 1: \(x \cdot \frac{1}{x} = 1\). This property is fundamental in understanding how numbers and their reciprocals behave when combined in equations or inequalities.

In our exercise, we examined the sum of a positive number \(x\) and its reciprocal, \(\frac{1}{x}\). We needed to prove that this sum can never be less than 2. The manipulation involved multiplying both sides of the starting inequality by \(x\) to eliminate the fraction, simplifying our calculations.

Understanding how reciprocals interact with other mathematical operations helps in solving a wide range of problems. When dealing with positive numbers, reciprocals ensure we are only juggling values that make logical sense and adhere to the inequality rules we seek to prove.

A perfect square is an expression of the form \((x - h)^2\) or \(x^2 - 2hx + h^2\), where \(h\) is a constant. Perfect squares are crucial in algebra because they are always non-negative. They reach their minimum value of zero at \(x = h\).

In this exercise, we turned the quadratic inequality \(x^2 - 2x + 1 \geq 0\) into a perfect square \((x - 1)^2 \geq 0\). Recognizing a perfect square simplifies proving the inequality, as it reveals the minimum value: zero.

• For \((x - 1)^2 \), the expression equals zero when \(x = 1\).
• Since squares are always non-negative, \((x - 1)^2\) cannot be less than zero.

This analysis confirms that the sum of a number and its reciprocal \(x + \frac{1}{x}\) has the minimum result of 2. Thus, using perfect squares helps effortlessly resolve the inequality we started with.