Một loại thuốc được dùng cho một bệnh nhân và nồng độ thuốc trong máu của bệnh nhân được giám sát bởi bác sĩ. Biết rằng nồng độ thuốc trong máu của bệnh nhân sau khi tiêm vào cơ thể trong t giờ được tính theo công thức \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4yamaabm % aabaGaamiDaaGaayjkaiaawMcaaiabg2da9maalaaabaGaamiDaaqa % aiaadshadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaIXaaaaaaa!3EF6! c\left( t \right) = \frac{t}{{{t^2} + 1}}\) (mg/L). Sau khi tiêm thuốc bao lâu thì nồng độ thuốc trong máu của bệnh nhân cao nhất?
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Lời giải:
Báo saiVới \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4yamaabm % aabaGaamiDaaGaayjkaiaawMcaaiabg2da9maalaaabaGaamiDaaqa % aiaadshadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaIXaaaaaaa!3EF6! c\left( t \right) = \frac{t}{{{t^2} + 1}}; t > 0\), ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabm4yayaafa % WaaeWaaeaacaWG0baacaGLOaGaayzkaaGaeyypa0ZaaSaaaeaacqGH % sislcaWG0bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGymaaqaam % aabmaabaGaamiDamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaigda % aiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaaaaaaa!44F1! c'\left( t \right) = \frac{{ - {t^2} + 1}}{{{{\left( {{t^2} + 1} \right)}^2}}}\).
Cho \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabm4yayaafa % WaaeWaaeaacaWG0baacaGLOaGaayzkaaGaeyypa0JaaGimaaaa!3B2A! c'\left( t \right) = 0\) \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aaS % aaaeaacqGHsislcaWG0bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIa % aGymaaqaamaabmaabaGaamiDamaaCaaaleqabaGaaGOmaaaakiabgU % caRiaaigdaaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaaaaOGa % eyypa0JaaGimaaaa!449B! \Leftrightarrow \frac{{ - {t^2} + 1}}{{{{\left( {{t^2} + 1} \right)}^2}}} = 0\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaam % iDaiabg2da9iaaigdaaaa!3B0A! \Leftrightarrow t = 1\)
Bảng biến thiên
Vậy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaaci % GGTbGaaiyyaiaacIhaaSqaamaabmaabaGaaGimaiaacUdacqGHRaWk % cqGHEisPaiaawIcacaGLPaaaaeqaaOGaam4yamaabmaabaGaamiDaa % GaayjkaiaawMcaaiabg2da9maalaaabaGaaGymaaqaaiaaikdaaaaa % aa!4457! \mathop {\max }\limits_{\left( {0; + \infty } \right)} c\left( t \right) = \frac{1}{2}\) khi t =1.