Biết rằng đồ thị hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadIhadaahaaWcbeqaaiaaiodaaaGccqGHRaWkcaaIZaGaamiE % amaaCaaaleqabaGaaGOmaaaaaaa!3D6E! y = {x^3} + 3{x^2}\) có dạng như hình vẽ:

Hỏi đồ thị hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maaemaabaGaamiEamaaCaaaleqabaGaaG4maaaakiabgUcaRiaa % iodacaWG4bWaaWbaaSqabeaacaaIYaaaaaGccaGLhWUaayjcSdaaaa!409A! y = \left| {{x^3} + 3{x^2}} \right|\) có bao nhiêu điểm cực trị?

Ta có:\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maaemaabaGaamiEamaaCaaaleqabaGaaG4maaaakiabgUcaRiaa % iodacaWG4bWaaWbaaSqabeaacaaIYaaaaaGccaGLhWUaayjcSdaaaa!409A! y = \left| {{x^3} + 3{x^2}} \right|\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0Zaai % qaaqaabeqaaiaadIhadaahaaWcbeqaaiaaiodaaaGccqGHRaWkcaaI % ZaGaamiEamaaCaaaleqabaGaaGOmaaaakiaaykW7caaMc8UaaGPaVl % aaykW7caaMc8UaaGPaVlaabUgacaqGObGaaeyAaiaaykW7caWG4bWa % aWbaaSqabeaacaaIZaaaaOGaey4kaSIaaG4maiaadIhadaahaaWcbe % qaaiaaikdaaaGccqGHLjYScaaIWaGaeyi1HSTaamiEaiabgwMiZkab % gkHiTiaaiodaaeaacqGHsislcaWG4bWaaWbaaSqabeaacaaIZaaaaO % GaeyOeI0IaaG4maiaadIhadaahaaWcbeqaaiaaikdaaaGccaaMc8Ua % aGPaVlaabccacaqGRbGaaeiAaiaabMgacaaMc8UaamiEamaaCaaale % qabaGaaG4maaaakiabgUcaRiaaiodacaWG4bWaaWbaaSqabeaacaaI % YaaaaOGaeyipaWJaaGimaiabgsDiBlaadIhacqGH8aapcqGHsislca % aIZaaaaiaawUhaaaaa!75BC! = \left\{ \begin{array}{l} {x^3} + 3{x^2}\,\,\,\,\,\,{\rm{khi}}\,{x^3} + 3{x^2} \ge 0 \Leftrightarrow x \ge - 3\\ - {x^3} - 3{x^2}\,\,{\rm{ khi}}\,{x^3} + 3{x^2} < 0 \Leftrightarrow x < - 3 \end{array} \right.\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0Zaai % qaaqaabeqaaiaadIhadaahaaWcbeqaaiaaiodaaaGccqGHRaWkcaaI % ZaGaamiEamaaCaaaleqabaGaaGOmaaaakiaaykW7caaMc8UaaGPaVl % aaykW7caaMc8UaaGPaVlaabUgacaqGObGaaeyAaiaaykW7caWG4bGa % eyyzImRaeyOeI0IaaG4maaqaaiabgkHiTiaadIhadaahaaWcbeqaai % aaiodaaaGccqGHsislcaaIZaGaamiEamaaCaaaleqabaGaaGOmaaaa % kiaaykW7caaMc8UaaeiiaiaabUgacaqGObGaaeyAaiaaykW7caWG4b % GaeyipaWJaeyOeI0IaaG4maaaacaGL7baaaaa!61C5! = \left\{ \begin{array}{l} {x^3} + 3{x^2}\,\,\,\,\,\,{\rm{khi}}\,x \ge - 3\\ - {x^3} - 3{x^2}\,\,{\rm{ khi}}\,x < - 3 \end{array} \right.\)

Nên ta lấy phần đối xứng của đồ thị hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadIhadaahaaWcbeqaaiaaiodaaaGccqGHRaWkcaaIZaGaamiE % amaaCaaaleqabaGaaGOmaaaaaaa!3D6E! y = {x^3} + 3{x^2}\) khi .x < -3

Dựa vào đồ thị, ta thấy đồ thị hàm số có 3 điểm cực trị.