My book asks me to show that $\tan x \gt x$ for $0\lt x\lt \pi/2$ by using the Mean-Value Theorem (MVT). The solution it gives is: $$\text{Let }f(x)=\tan x\text{.} \\ \text{If } 0\lt x\lt \pi/2\text{, then by the MVT } \\ f(x)-f(0)=f’(c)(x-0)\text{ for some } c \text{ in } (0,\pi/2) \text{.} \\ \text{Thus }\tan x = x\sec^2 c \gt x \text{, since }\sec c \gt 1\text{.}$$ My issue is the MVT says the function it applies to must be continuous on the closed, finite interval $[a, b]$, but $\tan$ is not continuous at $x=\pi/2$; even if the domain did incorporated it. Even worse, $x$ is restricted to $0\lt x$, so the function is not even continuous at the left endpoint of its domain. Why are we allowed to use the MVT here?

I’ve attempted to reformulate the theorem using an open interval, the problem is the proof relies on a result from Rolle’s Theorem, which is itself dependent on having a closed interval. In turn, Rolle’s Theorem relies on something called the Min-Max Theorem, which is not only dependent upon having a closed interval, but there is also no proof provided in my textbook, which states such a proof lies outside the scope of the book.

I am not learning analysis, I am only learning single variable calculus (currently differential calculus). If I am not able to, and I suspect not implicitly expected to, reformulate the theorems in order to make them applicable, then how is the MVT allowed to be used on this problem? The book also invokes it on a number of other instances where the functions in question are continuous only on an open interval. If a theorem is true of a closed interval, should I take it to mean that it is also applicable to an open interval, but not vice versa? That would make sense to me, if it was true in both directions it would be pointless to specify if a theorem is applicable to an open or closed interval, and it would explain why the book is justified in invoking it for functions continuous over open intervals.

  • 2
    $\begingroup$ Use some $b$ such that $x<b<\pi/2$ and apply the MVT to the interval $[0,b]$. $\endgroup$ – Jens Schwaiger Jun 24 at 3:28

The given solution is written a bit sloppy. Actually the mean value theorem is applied to the closed interval $[0, x]$ (on which $f$ is continuous and differentiable). The MVT states that $$ f(x)-f(0)=f’(c)(x-0)\text{ for some } c \text{ in } (0,x) \, , $$ and then the desired conclusion follows.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.