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?
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.
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.
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)