F(x) = x ^3 + 3 x ^2+ 4 x +b sin (x) + c cos(x) is one one function for all x belogs to R (real set).?

f(x) = x ^3 + 3 x ^2+ 4 x +b sin (x) + c cos(x) is one one function for all x belogs to R (real set). Then find the greatest value of





b^ 2 + c^ 2 ?

F(x) = x ^3 + 3 x ^2+ 4 x +b sin (x) + c cos(x) is one one function for all x belogs to R (real set).?
b² + c² is unbounded so it has neither local nor global extrema..








Doug
Reply:there is no greates value for f(x)


since the lim x-%26gt;infinity f(x) is infinity





now, b and c can be arbitrary, as large as you want or as small as you need.


so again, there is no gretes value of b^ 2 + c^ 2
Reply:Some progress...





Can rewrite as





f(x) = x³ + 3x² + 4x + √(b² + c²) sin ( x + atan(c/b) )





Is one-one if f '(x) ≥ 0 for all x





f '(x) = 3 (x+1)² + 1 + b cos x - c sin x.





Numerical experiments show that b = 13.31 and c = 20.74 is close to the maximum values of b and c, giving a maximum value of b² + c² of about 607.3 .





Continuing...





The condition on b and c: 3 (x+1)² + 1 + b cos x - c sin x ≥ 0 for all x defines a region in the b-c plane bounded by straight lines. The envelope of this family of lines is found from the above equation, and its derivative.





Changing b, c and x to x, y and t makes these equations more standard, where t is a parameter.





The family of lines is:





cos(t) x - sin(t) y + 3 (t +1)² + 1 = 0





The envelope curls around the origin and cuts itself at about (13, 21). This point of intersection is furtherest from the origin, corresponding to maximizing b² + c²





The derivative wrt t is:





-sin(t) x - cos(t) y + 6 (t+1) = 0





The point of self intersection corresponds to two parameters, s and t. The equations are





cos(t) x - sin(t) y + 3 (t +1)² + 1 = 0


-sin(t) x - cos(t) y + 6 (t+1) = 0





Solving for x and y:





x = 6 (t+1) sin t - [3 (t +1)² + 1] cos t





y = [3 (t +1)² + 1] sin t + 6 (t+1) cos t





Similarly for s, therefore:





6 (t+1) sin t - [3 (t +1)² + 1] cos t = 6 (s+1) sin s - [3 (s +1)² + 1] cos s





and





[3 (t +1)² + 1] sin t + 6 (t+1) cos t = [3 (s +1)² + 1] sin s + 6 (s+1) cos s





These 2 simultaneous equations in s and t can be solved using Newton's Method to get





s = -3.490373 and t = 1.490373





(Note that s + t = -2 !!! but why??)





Substituting, and using the original b and c instead of x and y, find that





b = 13.3188707 and c = 20.7429121 with b² + c² having a maximum value of





607.6607208.