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Mamont248 [21]
3 years ago
15

Let f(x)= [x/3] (where f(x) is the ceiling function). We learned that the floor and the ceiling functions are NOT invertible, bu

t we also learned about the set of preimages of any value in the Range, the set of images. Keeping that in mind, give your answer in interval notation if necessary.
a. Find f-1({5})
b. Find f-1({-2})
c. Find f-1({x | 5 = x = 9 })
d. Find f-1({x | -6 = x = -2})
Mathematics
1 answer:
meriva3 years ago
8 0

(a) We have ⌊<em>x</em>⌋ = 5 if 5 ≤ <em>x</em> < 6, and similarly ⌊<em>x</em>/3⌋ = 5 if

5 ≤ <em>x</em>/3 < 6   ==>   15 ≤ <em>x</em> < 18

(b) ⌊<em>x</em>⌋ = -2 if -2 ≤ <em>x</em> < -1, so ⌊<em>x</em>/3⌋ = -2 if

-2 ≤ <em>x</em>/3 < -1   ==>   -6 ≤ <em>x</em> < -3

In general, ⌊<em>x</em>⌋ = <em>n</em> if <em>n</em> ≤ <em>x</em> < <em>n</em> + 1, where <em>n</em> is any integer.

I do not understand what is being asked in (c) and (d), so you'll have to clarify...

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Find the values of x and y if (2x + y, 2) = (1, x - y). ​
m_a_m_a [10]

Answer:

We have, 2x+y=3 .......(1)

xy=1....(2)

In equation (2), we get

y=

x

1

..........(3)

y put in the equation (1)

2x+

x

1

=3

2x

2

+1=3x

2x

2

−3x+1=0

2x

2

−3x−1x+1=0

2x(x−1)−1(x−1)=0

x=1, x=

2

1

We take the value of x=1

x value put in equation (1)

2(1)+y=3

y=3−2

y=1

x=1

x and y value put (x+y)

x−y

=(1+1)

(1−1)

=2

0

=1

Hence, this is the answer

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6 0
2 years ago
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