Biết đồ thị hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadIhadaahaaWcbeqaaiaaiodaaaGccqGHsislcaaIZaGaamiE % aiabgUcaRiaaigdaaaa!3E2D! y = {x^3} - 3x + 1\) có hai điểm cực trị A, B. Khi đó phương trình đường thẳng AB là
Hãy suy nghĩ và trả lời câu hỏi trước khi xem đáp án
Lời giải:
Báo saiThực hiện phép chia y cho y' ta được: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iqadMhagaqbaiaac6cadaqadaqaamaalaaabaGaaGymaaqaaiaa % iodaaaGaamiEaaGaayjkaiaawMcaaiabgUcaRmaabmaabaGaeyOeI0 % IaaGOmaiaadIhacqGHRaWkcaaIXaaacaGLOaGaayzkaaaaaa!4470! y = y'.\left( {\frac{1}{3}x} \right) + \left( { - 2x + 1} \right)\)
Giả sử hai điểm cực trị của đồ thị hàm số lần lượt là: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqamaabm % aabaGaamiEamaaBaaaleaacaaIXaaabeaakiaacUdacaWG5bWaaSba % aSqaaiaaigdaaeqaaaGccaGLOaGaayzkaaaaaa!3CDF! A\left( {{x_1};{y_1}} \right)\) và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOqamaabm % aabaGaamiEamaaBaaaleaacaaIYaaabeaakiaacUdacaWG5bWaaSba % aSqaaiaaikdaaeqaaaGccaGLOaGaayzkaaaaaa!3CE2! B\left( {{x_2};{y_2}} \right)\).
Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadMhadaWgaaWcbaGaaGymaaqabaGccqGH9aqpcaWG5bWaaeWa % aeaacaWG4bWaaSbaaSqaaiaaigdaaeqaaaGccaGLOaGaayzkaaGaey % ypa0JabmyEayaafaWaaeWaaeaacaWG4bWaaSbaaSqaaiaaigdaaeqa % aaGccaGLOaGaayzkaaGaaiOlamaabmaabaWaaSaaaeaacaaIXaaaba % GaaG4maaaacaWG4bWaaSbaaSqaaiaaigdaaeqaaaGccaGLOaGaayzk % aaGaey4kaSYaaeWaaeaacqGHsislcaaIYaGaamiEamaaBaaaleaaca % aIXaaabeaakiabgUcaRiaaigdaaiaawIcacaGLPaaacqGH9aqpcqGH % sislcaaIYaGaamiEamaaBaaaleaacaaIXaaabeaakiabgUcaRiaaig % daaeaacaWG5bWaaSbaaSqaaiaaikdaaeqaaOGaeyypa0JaamyEamaa % bmaabaGaamiEamaaBaaaleaacaaIYaaabeaaaOGaayjkaiaawMcaai % abg2da9iqadMhagaqbamaabmaabaGaamiEamaaBaaaleaacaaIYaaa % beaaaOGaayjkaiaawMcaaiaac6cadaqadaqaamaalaaabaGaaGymaa % qaaiaaiodaaaGaamiEamaaBaaaleaacaaIYaaabeaaaOGaayjkaiaa % wMcaaiabgUcaRmaabmaabaGaeyOeI0IaaGOmaiaadIhadaWgaaWcba % GaaGOmaaqabaGccqGHRaWkcaaIXaaacaGLOaGaayzkaaGaeyypa0Ja % eyOeI0IaaGOmaiaadIhadaWgaaWcbaGaaGOmaaqabaGccqGHRaWkca % aIXaaaaiaawUhaaaaa!7810! \left\{ \begin{array}{l} {y_1} = y\left( {{x_1}} \right) = y'\left( {{x_1}} \right).\left( {\frac{1}{3}{x_1}} \right) + \left( { - 2{x_1} + 1} \right) = - 2{x_1} + 1\\ {y_2} = y\left( {{x_2}} \right) = y'\left( {{x_2}} \right).\left( {\frac{1}{3}{x_2}} \right) + \left( { - 2{x_2} + 1} \right) = - 2{x_2} + 1 \end{array} \right.\)
Ta thấy, toạ độ hai điểm cực trị A và B thoả mãn phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iabgkHiTiaaikdacaWG4bGaey4kaSIaaGymaaaa!3C3B! y = - 2x + 1\)
Vậy phương trình đường thẳng qua hai điểm cực trị là: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iabgkHiTiaaikdacaWG4bGaey4kaSIaaGymaaaa!3C3B! y = - 2x + 1\)