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Say I have a function $f\left(x\right)=\left|\frac{x^{2}-2}{x^{2}-1}\right|$. I have to find the local minima/maxima of the function. The point is that this function is non-differentiable at $x=\pm\sqrt{2},\pm1$. When i try to break free of the modulus, I end up getting a piece-wise function. How do I find the extrema points in this case ( I do not want to go by graph method) ? Because, sure I can find the critical points, which in this case is $x=\pm\sqrt{2},\pm1,\ 0$. But how do I use the first derivative test to check whether these points are actually extrema or not.

So basically my question is how do we go about finding local minima/maxima without any graphical method?

Any hints or approaches are welcome.

I don't want answer of this particular question, I just want to know the concept/method.

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  • $\begingroup$ You should expand $f(x)$ by the sign that it can have in certain intervals, and see $f(x)$ as a sectioned function and from there calculate the critical points $\endgroup$ Aug 6, 2021 at 11:54
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    $\begingroup$ Well, in general, if you have a function of the form $f(x) = |g(x)|$, then you can note that $f(x) \geqslant 0$ for all $x$ in the domain. Moreover, if there are points at which $g$ is zero, then they must be global minima for the function $f$. $\endgroup$ Aug 6, 2021 at 11:54
  • $\begingroup$ If you know a function $f: U \to \mathbb R$ for $U$ open has a singularity at a point $p \in \overline U$, then $\lim_{x \to p}|f(x)| = \infty$, This means that it is for certain that $f$ has either no global min or no global max. However, if you are only looking for local mins and max then one can use standard methodologies of finding critical points should $f$ be differentiable. $\endgroup$ Aug 6, 2021 at 11:54
  • $\begingroup$ @AryamanMaithani hmm, yes I understand. so in this case $\pm1$ is automatically canceled as $f\left(x\right)\ →\ ∞$ and $\pm\sqrt{2}$ are local minima because f(x)=0? $\endgroup$ Aug 6, 2021 at 12:01
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    $\begingroup$ @Vega: $\pm 1$ is cancelled because $f$ is not even defined at $\pm 1$. The part about $\pm \sqrt 2$ is correct. $\endgroup$ Aug 6, 2021 at 13:35

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Final Answer: (to summarize)

  1. In general, if we have a function of the form f(x)=|g(x)|, then you can note that f(x)⩾0 for all x in the domain. Moreover, if there are points at which g(x) is zero, then they must be global minima for the function f. (USEFUL POINT TO REMEMBER)

  2. If you know a function f:U→R for U open has a singularity at a point then limx→p|f(x)|=∞, This means that it is for certain that f has either no global min or no global max.

  3. If yet there is any other critical point, just use the fundamental definition that $f\left(c^{-}\right)>f\left(c\right)\ and\ f\left(c^{+}\right)>f\left(c\right)$ for minima and like that we can do for maxima. This can be used in a piece-wise function too at the point where the definition changes, to find if it is an extrema or not.

It is not a hard and fast method and depending upon the questions, things can change, nothing can be said exactly. Best way to solve these questions is the graphical approach wherever applicable.

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