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).?

Question

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 ?

Answer 1

b² + c² is unbounded so it has neither local nor global extrema..


Doug

Answer 2

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.

Answer 3

there is no greates value for f(x)
since the lim x->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