# Let F(s) = s+2/s+10. Find the maximum of F(s) on the interval [1,4].

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### 2 Answers

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.**