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Let $f(x)$ be continuous and differentiable on $[a,b]$. Show that if $f'(x)\leq 0$ for $x\in [a,\eta)$ and $f'(x)\geq 0$ for $x\in (\eta,b]$, then $f$ never takes a value smaller than $f(\eta)$.

$\textbf{First approach:}$ Since $f'(x)\leq 0$ for $x\in [a,\eta)$, then $f$ is decreasing on this interval. On the other hand, $f'(x)\geq 0$ for $x\in (\eta,b]$, then $f$ is increasing on this interval. According to this, $f(\eta)$ attains a minimum at $\eta$, and I just need to verify that $f'(\eta)=0$.\

$\textbf{Second approach:}$ Suppose that there exist $\eta_{0}$ such that $f(\eta_{0})< f(\eta)$ and continue to get a contradiction.

What approach should I follow?

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    $\begingroup$ @drhab: That is the “mean-value theorem.” $\endgroup$ – Martin R Jun 10 at 7:50
  • $\begingroup$ @MartinR Ah, yes. Thank you. My memory is not so good anymore :-). $\endgroup$ – drhab Jun 10 at 7:52
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Since $f$ is decreasing in $[a,\eta)$ and continuous it follows that $f(x) \geq f(\eta)$ in this interval. Similarly, $f(x) \geq f(\eta)$ in $[\eta,b]$ also. Hence $f(x) \geq f(\eta)$ for all $x$ which shows that we cannot have $f(x) <f(\eta)$ at any point.

[To see that $f(x) \geq f(\eta)$ for $x$ in $[a, \eta]$ note that $f(x) \geq f(\eta -\frac 1 n)$ (for $n$ so large that $\eta -\frac 1 n \in [a,\eta)$). Then let $n \to \infty$].

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If $x\in[a,b]-\{\eta\}$ then the mean value theorem tells us that:$$f(x)=f(\eta)+f'(\xi)(x-\eta)\tag1$$for some $\xi$ in the interval that has $x$ and $\eta$ as endpoints.

Now check that:

  • $f'(\xi)(x-\eta)\geq0$ if $x>\eta$ and consequently $f'(\xi)\geq0$.
  • $f'(\xi)(x-\eta)\geq0$ if $x<\eta$ and consequently $f'(\xi)\leq0$.

So according to $(1)$ we have $f(x)\geq f(\eta)$ for every $x\in[a,b]-\{\eta\}$.

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