# Analysis Function

• Nov 13th 2008, 05:51 PM
thahachaina
Analysis Function
Is there a function such that, f is continuous on (1,5), reaches a maximum on (1,5), but does not reach a minimum on (1,5). Does such a function exist or is it impossible?
• Nov 13th 2008, 06:34 PM
Rapha
Quote:

Originally Posted by thahachaina
Is there a function such that, f is continuous on (1,5), reaches a maximum on (1,5), but does not reach a minimum on (1,5). Does such a function exist or is it impossible?

f both reaches a maximum on (1,5) and a minimum on (1,5), never heard of that => i guess it reaches a max on (1,5), but not a minimum on (1,5) is possible.
• Nov 13th 2008, 07:17 PM
Mathstud28
Quote:

Originally Posted by thahachaina
Is there a function such that, f is continuous on (1,5), reaches a maximum on (1,5), but does not reach a minimum on (1,5). Does such a function exist or is it impossible?

Hint, basically think of a function that when differentiated gives a function such that $\displaystyle \exists{x_1}\in(1,5)\backepsilon{f(x_1)=0}$

This sounds hard, but it isn't. Think of trigonmetric function with an altered period for one.
• Nov 13th 2008, 09:34 PM
thahachaina
Thanks man. I kinda thought about it, but it looks to me like that function is not possible since it does not reach a maximum at (1,5). Is that right?
• Nov 13th 2008, 11:23 PM
CaptainBlack
Quote:

Originally Posted by thahachaina
Is there a function such that, f is continuous on (1,5), reaches a maximum on (1,5), but does not reach a minimum on (1,5). Does such a function exist or is it impossible?

$\displaystyle f(x)=-(x-1)(x-5)$

has a maximum at $\displaystyle x=3$ but has no minimum on the open interval $\displaystyle (1,5)$ .

(If it were a closed interval it would be a different matter)

CB