Because f(x+t)≥2 f(x) is always true, if the square of f(x)=x is brought in, the square of (X=t) ≥2X can be obtained. X for solving inequality is greater than or equal to t( 1- root number 2); Less than or equal to t( 1+ root number 2). And because x≥0, x∈[t, t+2], so t≥0, so the solution set of inequality x is x ≥ 0; Less than or equal to t( 1+ root number 2). Because x∈[t, t+2], X solution set "less than or equal to t( 1+ root number 2)" should be less than t+2, and the solution t is less than or equal to root number 2, and because t≥0, the value range of t is less than or equal to root number 2; Greater than or equal to 0