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Consider a function $y = x + 1/x$; $y_1=x$ is strictly increasing in $x \in (-\infty,\infty)$, and $y_2 = 1/x$ is strictly decreasing in $x \in (-\infty,0) \cup (0,\infty)$. The critical point do exist for both the solution space $x \in (-\infty,0)$ and $x \in (0,\infty)$.

Consider another function $y = x + 1/x^2$; $y_1=x$ is strictly increasing in $x \in (-\infty,\infty)$, and $y_2 = 1/x^2$ is strictly decreasing in $x \in (0,\infty)$. However, $y = 1/x^2$ is strictly increasing in $x \in (-\infty,0)$. The critical point exists for the solution space $x \in (0,\infty)$.

Consider a function $y = e^x + e^{-x}$; $y_1=e^x$ is strictly increasing in $x \in(-\infty,\infty)$, and $y_2 = e^{-x}$ is strictly decreasing in $x \in (-\infty,\infty)$. The critical point exists for the solution space $x \in (0,\infty)$.

As an extension, can we comment on the fact that if a function consists of one strictly increasing and one strictly decreasing component; and the function is a linear sum of its components there exist at least one critical point within the solution space, where the function is defined? If the statement is valid in general, can we prove it?

In addition, if strict condition is relaxed to monotone, will the conclusion still be valid?

What will be the effect of solution space on the existence of the critical point?

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    $\begingroup$ This isn't true. Consider $f(x) = 2x + (-x)$. $\endgroup$ – stochasticboy321 Sep 16 '16 at 4:30
  • $\begingroup$ Then, given a linear sum of two monotonic function, I can't comment on the existence of a critical point. Or can i? What if I change the statement to "there may exist"? In addition, I must look into case by case basis. $\endgroup$ – Sum Sep 16 '16 at 5:10
  • $\begingroup$ Well, but, 'there may exist a critical point' is a priori true for any differentiable function under the sun, right? $\endgroup$ – stochasticboy321 Sep 16 '16 at 6:25
  • $\begingroup$ Looking at your examples above, if $f$ and $g$ are the two functions, this might work if $fg \ge 0$ over the domain ($fg<0$ definitely doesn't work due to the previous counterexample). Perhaps give this a shot? (Fair warning, I'm about to crash, and this might be a sleep directed dead end). $\endgroup$ – stochasticboy321 Sep 16 '16 at 6:38
  • $\begingroup$ Not true.. If a function is a linear combination of two monotonic function, where both of them are simultaneously increasing or decreasing, in that case, I can guarantee that there does not exist exist any critical point (I am not talking about the saddle points). Correct me if I am wrong. Anyway, thank you vary much @stochasticboy321 $\endgroup$ – Sum Sep 16 '16 at 6:42

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