Question

函数f(x+1)=1+2t+2tx+x^2在区间（负无穷,3）上是减函数,则t的取值范围?

Answer 1

令m=x+1则x=m-1

f(m)=(m-1)²+2t(m-1)+2t+1

=m²-2m+1+2tm-2t+2t+1

=m²+(2t-1)m+2

f(x)=x²+(2t-1)x+2 (x≤2)

x=-(2t-1)/2≤2

t≥-3/2