已知f(x)的值域是【3/8,4/9】,g(x)=f(x)+√1-2f(x),试求y=g(x)的值域
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/10 07:52:55
已知f(x)的值域是【3/8,4/9】,g(x)=f(x)+√1-2f(x),试求y=g(x)的值域
已知f(x)的值域是【3/8,4/9】,g(x)=f(x)+√1-2f(x),试求y=g(x)的值域
已知f(x)的值域是【3/8,4/9】,g(x)=f(x)+√1-2f(x),试求y=g(x)的值域
因f(x)的值域是【3/8,4/9】,则√1-2f(x)的值域为【1/3,1/2】.令√1-2f(x)=a,则f(x)=1/2 (1- a^2),
g(x)=1/2 (1- a^2) a=-1/2 (a-1)^2 1
因为a的值域是【1/3,1/2】,g(x)在此值域内为增函数,所以g(x)的值域为【7/9,7/8】
3/8<=f(x)<=4/9
-8/9<=-2f(x)<=-3/4
1/9<=1-2f(x)<=1/4
1/3<=√[1-2f(x)]<=1/2
设t=√[1-2f(x)]
则f(x)=(1-t^2)/2
g(x)=(1-t^2)/2+t(1/3<=t<=1/2)
g(x)=-(1/2)t^2+t+1/2开口向下,对称轴为t=1
所以...
全部展开
3/8<=f(x)<=4/9
-8/9<=-2f(x)<=-3/4
1/9<=1-2f(x)<=1/4
1/3<=√[1-2f(x)]<=1/2
设t=√[1-2f(x)]
则f(x)=(1-t^2)/2
g(x)=(1-t^2)/2+t(1/3<=t<=1/2)
g(x)=-(1/2)t^2+t+1/2开口向下,对称轴为t=1
所以,在区间[1/3,1/2]递增。
最小值为-(1/2)*(1/3)+1/3+1/2=2/3
最大值为-(1/2)*(1/2)+1/2+1/2=3/4
y=g(x)的值域是[2/3,3/4]
收起
值域[49/81,23/32]
令m=f(x)
则3/8<=m<=4/9
g(x)=m+√(1-2m)
令a=√(1-2m)
3/8<=m<=4/9
1/9<=1-2m<=1/4
1/9<=a<=1/4
m=(1-a²)/2
所以y=g(x)=1/2-a²/2+a
=-1/2(a-1)²+1
1/9<=a<=1/4
所以是增函数
a=1/9,y=49/81
a=1/4,y=23/32
值域[49/81,23/32]