想問該怎麼求f(x)=-x^3-3x^2+6x+1的最大切線斜率? 感恩?
回答 (2)
2015-12-22 2:54 pm
y=－x^3－3x^2+6x+1
y’=－3x^2－6x+6
=－3(x^2+2x+1)+9
=－3(x+1)^2+9
(x+1)^2>=0
－3(x+1)^2<=0
－3(x+1)^2+9<=9
最大切線斜率=9
y’=－3x^2－6x+6
=－3(x^2+2x+1)+9
=－3(x+1)^2+9
(x+1)^2>=0
－3(x+1)^2<=0
－3(x+1)^2+9<=9
最大切線斜率=9
2015-12-20 11:18 pm
Sol
y=－x^3－3x^2+6x+1
y’=－3x^2－6x+6
=－3(x^2+2x+1)+9
=－3(x+1)^2+9
(x+1)^2>=0
－3(x+1)^2<=0
－3(x+1)^2+9<=9
最大切線斜率=9
y=－x^3－3x^2+6x+1
y’=－3x^2－6x+6
=－3(x^2+2x+1)+9
=－3(x+1)^2+9
(x+1)^2>=0
－3(x+1)^2<=0
－3(x+1)^2+9<=9
最大切線斜率=9
收錄日期: 2021-04-18 14:11:41
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20151220085513AAjytyz