Hi I am trying to understand and prove the fundamental theorem of calculus and I ran into some confusion understanding the intermediate value theorem . several sources online claim that if a function f(x) is continuous on [a,b] let s be a number such that f(a)<s<f(b) then there exists a number k in the open interval (a,b) such that f(k)=s my question is why do we only assume the open interval shouldn't it also include the closed interval [a,b] and also why does s have to be less than both f(a) and f(b)?
Calculus and the intermediate value theorem
calculusintegrationordinary differential equations
Best Answer
The assertion “there is some $k\in(a,b)$ such that $f(k)=s$” is stronger than the assertion “there is some $k\in[a,b]$ such that $f(k)=s$”. So, why would we state a weaker statement when we can as easiy prove a stronger one.
And, if you have in mind the statement “if $f(a)\leqslant s\leqslant f(b)$, then there is some $k\in[a,b]$ such that $f(k)=x$”, then that statement is trivial if $s=f(a)$ (just take $k=a$ then) or if $s=f(b)$ (just take $k=b$ then). So, the non-trivial part is when $f(a)<s<f(b)$.