F(s) = s + 2/s + 10. To find the maximum value of F(s) in the interval [1,4], we differentiate F(s) and equate the derivative to 0. Also, at the point of maximum value F''(s) is negative.
F(s) = s + 2/s + 10
F'(s) = 1 - 2/s^2
1 - 2/s^2 = 0
=> s^2 - 2 = 0
=> s = +sqrt 2 and s = -sqrt 2
But F''(s) = 6/s^3 which is positive, therefore there is no maximum value for F(s).
The required maximum value of F(s) does not exist.
First, we'll compute the critical points of the function. These points are the roots of the 1st derivative of the function.
We'll differentiate F(s).
F'(s) = (s+2/s+10)'
F'(s) = -2/s^2 We'll put -2/s^2 = 0 is not defined for any value of s.
We notice that the function F'(s) is negative and F(s) is decreasing for values of s in the range [1 ; 4] and F(s) has no maximum point.