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$$f(x)=\cot x-\sqrt 2 \csc x,\quad I=(0,\pi)$$

Show that the function $f$ has an absolute extremum on the given interval $I$ and find that value.

I've found the local maximum point from the first derivative. I've showed that the second derivative at that point is less than zero. I will find the values of endpoints but what points should I take as the endpoints on an open interval? I'll also find the value of the critical point and say that is the only critical point on the interval and then compare all values to determine which one the absolute maximum value is. Are these steps right?

Kamil Jarosz
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owlswollen
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1 Answers1

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Sounds about right! To find the absolute extrema of a differentiable (!) function on an interval, one should indeed check the critical points (where the first derivative is zero) and the boundary points, then compare all found values and pick the largest (smallest).

As you pointed out, sometimes intervals are open and boundaries may or may not be $+\infty$ of $-\infty$. In this case, you should resort to using limits instead of just evaluating in the boundary points. In you case, calculate $$ \lim_{x\to 0^{+}} f(x) \quad \text{ and } \quad \lim_{x \to \pi^{-}} f(x) $$ If one of these values (which may be infinite) is larger (smaller) than the values attained in the critical points, than your function has an extremum at the boundaries. If the boundaries are not included in your domain, your function then does not have an absolute maximum (minimum) within the given interval.

Let me know if anything is still unclear.

Bib-lost
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