Extra properties of the intermediate value theorem on a differentiable interval

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

• 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. Jan 16, 2022 at 12:49
• @DavidMitra Thanks, how to prove there is a largest c in [a,b) ? Jan 18, 2022 at 11:00
• If not, then, using continuity of $f$ at $b$, we'd have $f(b)=y_0$. Jan 18, 2022 at 11:05
• @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? Jan 22, 2022 at 15:19
• There is a sequence $(x_n)$ in $C$ converging to $M$. As $f$ is continuous, $f(M)=\lim f(x_n)$. Jan 22, 2022 at 15:23

If $$f(x)$$ is continuous over the closed interval $$[a, b]$$ and differentiable at every point of its interior $$(a, b)$$ with $$f(a), and if $$y_{0}$$ is any value strictly between $$f(a)$$ and $$f(b)$$, that is $$f(a), 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 $$M-\varepsilon_{n}, so that one can find a sequence $$\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), 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 $$f(d) $$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.