f(x)=log^2(x+1)/(x-1)+log^2(x-1)+log^2(p-x)的值域(负无穷,log^(p+1)
来源:学生作业帮 编辑:搜搜考试网作业帮 分类:数学作业 时间:2024/05/28 01:22:43
f(x)=log^2(x+1)/(x-1)+log^2(x-1)+log^2(p-x)的值域(负无穷,log^(p+1)^2/4],则p的取值范围是?
对称轴应该大于1吧?要不然取不到最大值,p/2大于1,所以p应该大于3吧
对称轴应该大于1吧?要不然取不到最大值,p/2大于1,所以p应该大于3吧
f(x)=log^2(x+1)/(x-1)+log^2(x-1)+log^2(p-x)
定义域:(x+1)/(x-1)>0,且x-1>0,且p-x>0
x<-1或x>1,且x>1,且x<p
∴1<x<p
f(x)=log2 (x+1)/(x-1)+log2 (x-1)+log2 (p-x)
=log2 [(x+1)/(x-1) * (x-1) * (p-x) ]
=log2 [(x+1) (p-x) ]
=log2[-x^2+(p-1)x+p]
令g(x)=-x^2+(p-1)x+p,其最大值:[4*(-1)*p-(p-1)^2]/[4*(-1)]=(p+1)^2/4恒大于0
另外,对称轴x=(p-1)/2应符合定义域1<x<p的要求:
1<(p-1)/2<p
2<p<2p+1
故P>3
定义域:(x+1)/(x-1)>0,且x-1>0,且p-x>0
x<-1或x>1,且x>1,且x<p
∴1<x<p
f(x)=log2 (x+1)/(x-1)+log2 (x-1)+log2 (p-x)
=log2 [(x+1)/(x-1) * (x-1) * (p-x) ]
=log2 [(x+1) (p-x) ]
=log2[-x^2+(p-1)x+p]
令g(x)=-x^2+(p-1)x+p,其最大值:[4*(-1)*p-(p-1)^2]/[4*(-1)]=(p+1)^2/4恒大于0
另外,对称轴x=(p-1)/2应符合定义域1<x<p的要求:
1<(p-1)/2<p
2<p<2p+1
故P>3
f(x)=log^2(x+1)/(x-1)+log^2(x-1)+log^2(p-x)的值域(负无穷,log^(p+1)
log(X+5)+log(X+2)=1
f(x)=|log(a)(x)-1|+|2log(a)(x)|,求使f(x)<2的x范围,
lg(7*2^x+8)>=log(√10) 2^x,求函数f(x)=log(1/2)x*log(1/2)x/4的最小值
1/log(2)x+1/log(3)x+1/log(4)x=1,x=?
已知log(1/7)[log(3)(log(2)x)]=0
log a(1/2次方) x等于2log a
(高一数学)y=log(2)2x-log(1/2)x的三次方-4的值域
函数f(x)=log(2)(1+x)定义域
log底数7[log底数3(log底数2X)]=0,那么x^1/2的值是
log(0.1)log(0.2)log(0.5)(1/根号5次2)=x,求x的值
(高一)若x满足2(log(1/2)x)^2-14log(4)x+3≤0,求f(x)=[log(2)(x/2)]*{lo