f(x)=kx*x+(1-2k)x+1有最大值4,求k值f(x)=kx*x+(1-2k)x+1在[1,3]有最大值4求k值 忘打了抱歉

来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/29 02:59:33
f(x)=kx*x+(1-2k)x+1有最大值4,求k值f(x)=kx*x+(1-2k)x+1在[1,3]有最大值4求k值忘打了抱歉f(x)=kx*x+(1-2k)x+1有最大值4,求k值f(x)=k

f(x)=kx*x+(1-2k)x+1有最大值4,求k值f(x)=kx*x+(1-2k)x+1在[1,3]有最大值4求k值 忘打了抱歉
f(x)=kx*x+(1-2k)x+1有最大值4,求k值
f(x)=kx*x+(1-2k)x+1在[1,3]有最大值4求k值 忘打了抱歉

f(x)=kx*x+(1-2k)x+1有最大值4,求k值f(x)=kx*x+(1-2k)x+1在[1,3]有最大值4求k值 忘打了抱歉
f(x)=kx^2+(1-2k)+1
有最大值
k<0
f(x)=k[x+(1-2k)/2k]^2+1-(1-2k)^2/4k
x=(1-2k)/2k时,f(x)最大值=4
1-(1-2k)^2/4k=4
(1-2k)^2=-12k
4k^2+8k+1=0
4(k+1)^2=3
k=-√3/2-1 [ k=1-√3/2>0舍去】

顶点纵坐标为:[4k-(1-2k)^2]/4k=4, 由最大值可知开口向下,k<0
4k^2+8k+1=0
k=-1±√3/2

[4k-(1-2k)²]/(4k)=4,(k<0)

k=-√3/2-1 [ k=1-√3/2>0舍去】