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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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4 0
4 years ago
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3 years ago
HELP MATH ASAP! MARKING BRANLIST
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8 0
3 years ago
Compare the values of each digit. 6,237.2 . The 2 in the hundreds place is _____the value of the 2 in the tenths place?
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4 0
4 years ago
Please help, image is attached! Thanks.
Mkey [24]

When something is bisected, it is cut into two equal halves.

We know that SV cut RST into two parts.

Those two parts are RSV and VST

RSV = 64 and VST = 64

RST = RSV + VST = 64 + 64 = 128

The correct answer should be 128

5 0
3 years ago
Read 2 more answers
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