Question

F(s)=(s+2)/(s+9)
find the max of F(s) on interval [1,4]

Answer 1

quotient rule first?

Answer 2

\[F'(s)=\frac{d}{ds}\frac{s+2}{s+9}=\frac{(1)(s+9)-(s+2)(1)}{(s+9)^2}=0 \implies \frac{7}{(s+9)^2}=0\]
This is never true for any s. This means there are no local max/min. So test the end points:
\[f(1)=\frac{3}{10}, f(4)=\frac{6}{13}\]
So 6/13 should be the max. and 3/10 should be the min.

Answer 3

THANK YOU