f(x)=1/3x³-(a+1)x²/2 +a,当a>0时,求f(x)零点的个数
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f(x)=1/3x³-(a+1)x²/2 +a,当a>0时,求f(x)零点的个数
f(x)=1/3x³-(a+1)x²/2 +a,当a>0时,求f(x)零点的个数
f(x)=1/3x³-(a+1)x²/2 +a,当a>0时,求f(x)零点的个数
数理答疑团为您解答,希望对你有所帮助.
1/3x³-(a+1)x²/2 +a=0
2x³-3(a+1)x² +6a=0
a>0时,方程可有3个解
故:f(x)=1/3x³-(a+1)x²/2 +a,当a>0时,求f(x)零点的个数为3.
更上一层楼!
先运用导函数F'(x)=X^2-(a+1)X=X*(1+a);再令F(X)=0,求出X的值;分类讨论画出原函数的图像即可知道