التفكير الناقد وتخطيط المنحنيات

Critical Points and Curve Sketching

• Critical points are found by setting the derivative of a function equal to zero.
• The given function is y = x³ - 3x² - 9x.
• The derivative is y' = 3x² - 6x - 9.
• Setting y' = 0 gives 3x² - 6x - 9 = 0.
• Simplifying the equation gives x² - 2x - 3 = 0.
• Factoring the equation gives (x - 3)(x + 1) = 0.
• The solutions are x = 3 and x = -1.
• These are the critical points.

Calculating y-values

• Substituting x = 3 into the original function gives y = (3)³ - 3(3)² - 9(3) = 27 - 27 - 27 = -27.
• Substituting x = -1 into the original function gives y = (-1)³ - 3(-1)² - 9(-1) = -1 - 3 + 9 = 5.
• The critical points are (3, -27) and (-1, 5).

Increasing/Decreasing Intervals

• To determine where the function is increasing or decreasing, test values in the intervals defined by the critical points.
• Test values from (-∞, -1): Choose x = -2. y' = 3(-2)² - 6(-2) - 9 = 12 + 12 - 9 = 15 > 0, so the function is increasing.
• Test values from (-1, 3): Choose x = 0. y' = 3(0)² - 6(0) - 9 = -9 < 0, so the function is decreasing.
• Test values from (3, ∞): Choose x = 4. y' = 3(4)² - 6(4) - 9 = 48 - 24 - 9 = 15 > 0, so the function is increasing.

Summary

• The function increases on (-∞, -1) and (3, ∞).
• The function decreases on (-1, 3).

Additional Information (from the image)

• The y-values are calculated at x = 1 and x = 2, providing additional points to help understand the shape of the curve.
• A number line diagram was used to represent the intervals of increasing and decreasing behavior.

هذا الاختبار يتناول النقاط الحرجة وتخطيط المنحنيات لدالة رياضية معينة. سيتم استكشاف كيفية تحديد النقاط الحرجة من خلال الاشتقاق وكيفية حساب القيم المقابلة لها. ينطوي الاختبار على معرفة كيفية تحديد الفترات المتزايدة والمتناقصة للدالة.