The second group is substituted into the formula h(x)=a*f 1(x)+b*f2(x), and the comparison coefficient method shows that it does not meet the requirements.
Secondly, the objective function is regarded as a linear function about t, which shows that the objective function increases monotonously when x∈, so when the value x=2 is substituted into the function, t
The coordinate of the third question is the lowest, and the sum function h(x)=ax+b/x≥2 ab, and the solution is A = 2 and B = 8. Solve h(x
), substitute h(x 1)h(x2) and expand it, and establish the condition of x 1+x2= 1. The minimum value of g(x)=h(x 1)h(x2) can be obtained from the monotonicity of the function, that is, the maximum constant m.
Thirdly, because A > 0, B > 0, and h(X)=ax+b/x≥2 root signs, ab is equal to 8 and X 2 is equal to 2 2 = B/A, and these two equations are solved simultaneously, and A = is substituted to get H (x).
Because x1x 2 = x1(1-x1) =-(x1-1/2) 2+1/4.
From the monotonicity of function h(x) (when the two functions are viewed separately), we can know that 4 (x1x 2+16/x1x 2) is decreasing on (0, 1/4), (/kloc-) 1 6 (x 1/x2+x2/x1) decreases at (0,1) and increases at (1,positive infinity), so the function is at x1x2 =. At this time, x1x2 =16/x1x2 and x 1/x2=x2/x 1, that is, x 1=x2= 1/2, and substitute for H. =m is a constant, so m should be less than or equal to the minimum value of the function, that is, the maximum constant m=289 if and only if x 1=x2= 1/2.
Ps: It's really inconvenient to explain on the computer. ...