Here's a statement of the EVT: Let #f# be continuous on #[a,b]#. Then there exist numbers #c,d\in [a,b]# such that #f(c)\leq f(x)\leq f(d)# for all #x\in [a,b]#. Stated another way, the "supremum " #M# and "infimum " #m# of the range #\{f(x):x\in [a,b]\}# exist (they're finite) and there exist numbers #c,d\in [a,b]# such that #f(c)=m# and #f(d)=M#.
Note that the function #f# must be continuous on #[a,b]# for the conclusion to hold. For example, if #f# is a function such that #f(0)=0.5#, #f(x)=x# for #0<x<1#, and #f(1)=0.5#, then #f# attains no maximum or minimum value on #[0,1]#. (The supremum and infimum of the range exist (they're 1 and 0, respectively), but the function never attains (never equals) these values.)
Note also that the interval must be closed. The function #f(x)=x# attains no maximum or minimum value on the open interval #(0,1)#. (Once again, the supremum and infimum of the range exist (they're 1 and 0, respectively), but the function never attains (never equals) these values.)
The function #f(x)=1/x# also does not attain a maximum or minimum value on the open interval #(0,1)#. Moreover, the supremum of the range does not even exist as a finite number (it's "infinity").
Here's a statement of the IVT: Let #f# be continuous on #[a,b]# and suppose #f(a)!=f(b)#. If #v# is any number between #f(a)# and #f(b)#, then there exists a number #c\in (a,b)# such that #f(c)=v#. Moreover, if #v# is a number between the supremum and infimum of the range #{f(x):x\in [a,b]}#, then there exists a number #c\in [a,b]# such that #f(c)=v#.
If you draw pictures of various discontinuous functions, it's pretty clear why #f# needs to be continuous for the IVT to be true.
