Question

Let
F(s)=(s+5)/(s+11)
Find the maximum of F(s) on the interval [1,8].
I got my derivative as f'(s)=(6)/(s+11)^2. Then I set it to zero and I get no solution. Help?

Answer 1

that's ok. if there's no zero in the interval then evaluate the function at the end points of the interval and compare.

Answer 2

\[F(s) = \frac{ s+5 }{ s+11 } \Rightarrow F'(s) = \frac{ (s+11)-(s+5) }{(s+11)^2 }= \frac{ 6 }{ (s+11)^2 }\Rightarrow F'(s) >0, \forall \{s \in (1 \le s \le 8)\}\]

Answer 3

so too find the max, evaluate at the end points. since F'(s) is increasing in the interval, the max should be at s = 8.

Answer 4

\[F'(s) > 0, \forall \{s \in (1 \le s \le 8)\}\]

Answer 5

So instead of setting it equal to zero it would be 8?

Answer 6

no, since you can't make the derivative 0, you have to evaluate the original function at the endpoints.

find F(1) and F(8) and compare to see which is bigger. The bigger one will be the max.

this is a result of the extreme value theorem.

Answer 7

Thank you!

Answer 8

you're welcome