作业帮已知函数f(x)=log1/2(x+1),当点P（xo,yo)在y=f(x)的图象上移动时,点Q[(xo-t+10/2,yo](t∈R)在函数y=g(x)已知函数f(x)=log1/2(x+1),当点P（xo,yo)在y=f(x)的图象上移动时，点Q[(xo-t+1)/2,yo](t∈R)在函数y=g(x)的图象上
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在函数y=g(x)已知函数f(x)=log1/2(x+1),当点P（xo,yo)在y=f(x)的图象上移动时，点Q[(xo-t+1)/2,yo](t∈R)在函数y=g(x)的图象上](/uploads/image/z/10096293-21-3.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dlog1%2F2%28x%2B1%29%2C%E5%BD%93%E7%82%B9P%EF%BC%88xo%2Cyo%29%E5%9C%A8y%3Df%28x%29%E7%9A%84%E5%9B%BE%E8%B1%A1%E4%B8%8A%E7%A7%BB%E5%8A%A8%E6%97%B6%2C%E7%82%B9Q%5B%28xo-t%2B10%2F2%2Cyo%5D%28t%E2%88%88R%29%E5%9C%A8%E5%87%BD%E6%95%B0y%3Dg%28x%29%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dlog1%2F2%28x%2B1%29%2C%E5%BD%93%E7%82%B9P%EF%BC%88xo%2Cyo%29%E5%9C%A8y%3Df%28x%29%E7%9A%84%E5%9B%BE%E8%B1%A1%E4%B8%8A%E7%A7%BB%E5%8A%A8%E6%97%B6%EF%BC%8C%E7%82%B9Q%5B%28xo-t%2B1%29%2F2%2Cyo%5D%28t%E2%88%88R%29%E5%9C%A8%E5%87%BD%E6%95%B0y%3Dg%28x%29%E7%9A%84%E5%9B%BE%E8%B1%A1%E4%B8%8A)
已知函数f(x)=log1/2(x+1),当点P（xo,yo)在y=f(x)的图象上移动时,点Q[(xo-t+10/2,yo](t∈R)在函数y=g(x)已知函数f(x)=log1/2(x+1),当点P（xo,yo)在y=f(x)的图象上移动时，点Q[(xo-t+1)/2,yo](t∈R)在函数y=g(x)的图象上
已知函数f(x)=log1/2(x+1),当点P（xo,yo)在y=f(x)的图象上移动时,点Q[(xo-t+10/2,yo](t∈R)在函数y=g(x)
已知函数f(x)=log1/2(x+1),当点P（xo,yo)在y=f(x)的图象上移动时，点Q[(xo-t+1)/2,yo](t∈R)在函数y=g(x)的图象上移动。
①若点P坐标为（1，-1），点Q也在y=f(x)的图象上，求t的值；
②求函数y=g(x)的解析式；
③当t＞0时，试探求一个函数h(x)，使得f(x)+g(x)+h(x)在限定定义域为[0，1）时有最小值而没有最大值。
已知函数f(x)=log1/2(x+1),当点P（xo,yo)在y=f(x)的图象上移动时,点Q[(xo-t+10/2,yo](t∈R)在函数y=g(x)已知函数f(x)=log1/2(x+1),当点P（xo,yo)在y=f(x)的图象上移动时，点Q[(xo-t+1)/2,yo](t∈R)在函数y=g(x)的图象上
Ⅰ.P(1,-1)=>x0=1,y0=-1=>Q(2/3-t,-1)打入计算-1=log1/2(2/3-t+1) =>t=2/1;
Ⅱ.令X=Xo-t+1/2,Y=Yo =>Xo=X+t-2/1,Yo=Y带入F(X)可得：Y=log1/2(X+t-1/2+1) =>
G(x)=log1/2(x+t+1/2);
Ⅲ.h(x)=log1/2(1-x)
（1）t=2
(2)X=(X0-t+1)/2 Y=Y0,得X0=2X+t-1,Y0=Y
（xo,yo)在y=f(x)的图象上,g(x)=log1/2(2X+t-1)