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The intermediate value theorem for continuous functions:

If $f$ is a continuous function on a closed interval $[a, b]$, and if $y_{0}$ is any value between $f(a)$ and $f(b)$(that is $f(a)<y_{0}<f(b)$ or $f(a)>y_{0}>f(b)$), then $y_{0}=f(c)$ for some $c$ in $(a, b)$.

If $f$ is differentiable on (a, b) and $f(a)<f(b)$, is there such $c$ so that $y_{0}=f(c)$ for some $c$ in $(a, b)$ and $f'(c) \geq 0$?

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    $\begingroup$ You could show there is a largest $c$ in $[a,b)$ with $f(c)=y_0$ and then argue that the derivative can not be negative there. $\endgroup$ Jan 16, 2022 at 12:49
  • $\begingroup$ @DavidMitra Thanks, how to prove there is a largest c in [a,b) ? $\endgroup$
    – iMath
    Jan 18, 2022 at 11:00
  • $\begingroup$ If not, then, using continuity of $f$ at $b$, we'd have $f(b)=y_0$. $\endgroup$ Jan 18, 2022 at 11:05
  • $\begingroup$ @DavidMitra Thanks for the tip, but I can only show the set $C=\left\{c \mid f(c)=y_{0}\right\}$ has a least upper bound M, how to prove M is a member of C? $\endgroup$
    – iMath
    Jan 22, 2022 at 15:19
  • $\begingroup$ There is a sequence $(x_n)$ in $C$ converging to $M$. As $f$ is continuous, $f(M)=\lim f(x_n)$. $\endgroup$ Jan 22, 2022 at 15:23

1 Answer 1

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If $f(x)$ is continuous over the closed interval $[a, b]$ and differentiable at every point of its interior $(a, b)$ with $f(a)<f(b)$, and if $y_{0}$ is any value strictly between $f(a)$ and $f(b)$, that is $f(a)<y_{0}<f(b)$, then there is some $c$ in $(a, b)$ such that $y_{0}=f(c)$ and $f^{\prime}(c) \geq 0$, and even further the satisfying $c$ could be the greatest or the least number in set $\left\{c \mid f(c)=y_{0}\right.$ and $c$ in $\left.(a, b)\right\}$.

The proof as follows. According to the intermediate value theorem, there is a set $C=\left\{c \mid f(c)=y_{0}\right.$ and $c$ in $\left.(a, b)\right\}$, since $b$ is an upper bound for $C$, there should be a least upper bound $M$ for $C$. if $C$ is a finite set, then $M \in C$, while if $C$ is an infinite set, then for every positive real number $\varepsilon_{n}$, there is a real number $c_{n}$ in $C$ such that $M-\varepsilon_{n}<c_{n} \leq M$, so that one can find a sequence $\left\{c_{n}\right\}$ with the limit $M$, because of the continuity of $f(x)$, therefore $\lim _{n \rightarrow \infty} f\left(c_{n}\right)=\lim _{n \rightarrow \infty} y_{0}=y_{0}=f(M)$, thus $M$ is the greatest one in $C$ whether $C$ is infinite or not. Now supposing $f^{\prime}(M)<0$ in case of $f(a)<f(b)$, that is $f(x)$ is decreasing at $M$, then there is a $d$ in some right neighborhood of $M$ such that $f(M)=y_{0}>f(d)$, then according to the intermediate value theorem there should be another $y_{0}$ in the range of $f$ such that $f(d)<y_{0}<$ $f(b)$ on $(d, b)$, which is contradictory to the definition of $M$, thus $f^{\prime}(M) \geq 0$. Similar arguments for the lower bound $a$ and the case $f(a)>f(b)$ totally complete the proof.

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