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Usimov [2.4K]
2 years ago
12

Over the closed interval [3, 8], for which function can the extreme value theorem be applied? h (x) = startfraction negative 2 o

ver 5 (x minus 4) squared endfraction h (x) = startfraction (x minus 5) (x minus 1) over x squared minus 25 endfraction h (x) = startlayout enlarged left-brace first row startfraction 9 x over 10 minus x endfraction, x less-than 4 second row x 2, x greater-than-or-equal-to 4 endlayout h (x) = startlayout enlarged left-brace first row negative x, x less-than 5 second row x squared minus 20, x greater-than-or-equal-to 5 endlayout
Mathematics
1 answer:
Aleks04 [339]2 years ago
7 0

Over the closed interval [3, 8], the function that the extreme value theorem be applied is the third option.

<h3>What is the extreme value theorem?</h3>

The extreme value theorem can be applied when the function is continuous in the interval. Here, the function is continuous when the points in the interval are in the domain of the function.

Therefore, over the closed interval [3, 8], the function that the extreme value theorem be applied is the third option. This is because the function is continuous over the interval.

Learn more about theorem on:

brainly.com/question/23405626

#SPJ4

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