当a为何值时,函数f(x)=1\3x^3+(a+2)x^2 +(2a+1)x+1没极值点.
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/04 05:17:19
![当a为何值时,函数f(x)=1\3x^3+(a+2)x^2 +(2a+1)x+1没极值点.](/uploads/image/z/14305557-21-7.jpg?t=%E5%BD%93a%E4%B8%BA%E4%BD%95%E5%80%BC%E6%97%B6%2C%E5%87%BD%E6%95%B0f%28x%29%3D1%5C3x%5E3%2B%28a%2B2%29x%5E2+%2B%282a%2B1%29x%2B1%E6%B2%A1%E6%9E%81%E5%80%BC%E7%82%B9.)
当a为何值时,函数f(x)=1\3x^3+(a+2)x^2 +(2a+1)x+1没极值点.
当a为何值时,函数f(x)=1\3x^3+(a+2)x^2 +(2a+1)x+1没极值点.
当a为何值时,函数f(x)=1\3x^3+(a+2)x^2 +(2a+1)x+1没极值点.
f(x)=1\3x^3+(a+2)x^2 +(2a+1)x+1
求导,得
f'(x)=x^2+2(a+2)x +2a+1
△=4(a+2)^2-4(2a+1)=4(a^2+4a+4)-8a-4=4a^2+8a+12=4(a+1)^2+8>0
因为没有极值点,所以此时△应
f(x)=1/3x^3+(a+2)x^2 +(2a+1)x+1
f'(x)=x^2+2(a+2)x+(2a+1)
没极点即f'(x)=0无解或只有一解
△=4(a+2)^2-4(2a+1)<=0
得a的范围是[-3,2]