The greatest integer function returns the largest integer smaller than the number you provide it. That is, if n ≤ x < n + 1, where n is an integer, then the "greatest integer of x" is [x] = n.

• Let n be even. Then we can write n = 2k for some integer k ≥ 0. Now,

[n/2] = [k] = k

while

[-n/2] = [-k] = -k

so that

[n/2] - [-n/2] = 2k = n

• Let n be odd. Then n = 2k + 1 for some integer k ≥ 0. Every odd integer occurs between two even integers, so that

n - 1 < n < n + 1

or equivalently,

2k < n < 2k + 2

so that

k < n/2 < k + 1

It follows that [n/2] = k.

Similarly, if we negative the previous inequality, we have

-k > -n/2 > -(k + 1), or -k - 1 < -n/2 < -k

which means [-n/2] = -k - 1.

So we make the same conclusion,

[n/2] - [-n/2] = k - (-k - 1) = 2k + 1 = n

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2 years ago

The teacher kept track of how long it took students to finish a test. After 20 minutes, 3 students had finished the test.

Olin [163]

9 students finished the test

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I need to know the answer plz

sergij07 [2.7K]

Answer:

the answer is c.

Step-by-step explanation:

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Explain why Rolle's Theorem does not apply to the function even though there exist a and b such that f(a) = f(b). f(x) = cot x/

nydimaria [60]

$cot(x/2)$ is not differentiable over the interval $[ \pi, 7\pi]$

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