\(x \in (1, 3)\) の場合、\(x = 2\) をとぶ: \(3(2)^2 - 12(2) + 9 = -3 \leq 0\)。

["Title: Understanding the Inequality (3x^2 - 12x + 9 \leq 0) Over the Interval (x \in (1, 3))", "---", "When exploring quadratic inequalities, one key task is determining the values of (x) that satisfy expressions over specific intervals. A compelling example is analyzing whether (x = 2) lies within the solution set of the inequality:", "[\n3x^2 - 12x + 9 \leq 0 \quad \ ext{for} \quad x \in (1, 3).\n]", "In this article, we’ll break down the inequality, verify the result at (x = 2), and explain how (x = 2) satisfies the condition.", "---", "### What is the Inequality (3x^2 - 12x + 9 \leq 0)?", "We start by examining the quadratic expression:", "[\nf(x) = 3x^2 - 12x + 9.\n]", "This is a parabola opening upwards (since the coefficient of (x^2) is positive). To solve (3x^2 - 12x + 9 \leq 0), we first find the roots of the corresponding equation:", "[\n3x^2 - 12x + 9 = 0.\n]", "Dividing through by 3 simplifies:", "[\nx^2 - 4x + 3 = 0.\n]", "Factoring gives:", "[\n(x - 1)(x - 3) = 0,\n]", "so the roots are (x = 1) and (x = 3). These divide the real line into intervals:", "- (x < 1)\n- (1 \leq x \leq 3)\n- (x > 3)", "We now determine the sign of (f(x)) in each interval:\n- For (x < 1): test (x = 0): (3(0)^2 - 12(0) + 9 = 9 > 0).\n- For (1 < x < 3): test (x = 2): (3(4) - 24 + 9 = 12 - 24 + 9 = -3 \leq 0).\n- For (x > 3): test (x = 4): (3(16) - 48 + 9 = 48 - 48 + 9 = 9 > 0).", "Thus, the inequality (3x^2 - 12x + 9 \leq 0) holds only on the closed interval:", "[\nx \in [1, 3].\n]", "---", "### Checking if (x = 2) Belongs to the Solution Set", "Since (x = 2) lies in the interval (1 < x < 3), and we showed the expression is (\leq 0) exactly on ([1, 3]), it follows that:", "[\nf(2) = 3(2)^2 - 12(2) + 9 = 12 - 24 + 9 = -3 \leq 0.\n]", "Hence, (x = 2) satisfies the inequality. Moreover, because (2 \in (1, 3)), the solution meets both the inequality requirement and the domain constraint.", "---", "### Why Does This Matter?", "Understanding such intervals helps in countless real-world applications—from physics to optimization—where constraints define allowed values. In this case, the interval (x \in (1, 3)) represents a bounded domain, and the inequality (f(x) \leq 0) identifies feasible points where the quadratic function behaves as expected.", "---", "### Summary", "- The inequality (3x^2 - 12x + 9 \leq 0) holds if and only if (x \in [1, 3]).\n- Since (1 < 2 < 3), (x = 2) is within the solution set.\n- Verification confirms: (f(2) = -3 \leq 0).\n- Thus, the statement “(x = 2) is taken” (within the interval and satisfying the inequality) is true.", "---", "Key Takeaway:\nAlways test critical points and endpoints in quadratic inequalities over bounded intervals. For the expression (3x^2 - 12x + 9), (x = 2) lies in the valid range ([1, 3]) and satisfies the condition, making it a valid solution.", "---", "Related Searches:\n- Solve quadratic inequalities over an interval\n- Test values in (x^2) inequality\n- Find where (3x^2 - 12x + 9 = 0)\n- Evaluate polynomial expressions in domain restrictions", "---", "Keywords: (x \in (1, 3)), (3x^2 - 12x + 9 \leq 0), inequality solution, test (x = 2), quadratic functions, domain constraints, educational math, algebra practice."]