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Molodets [167]
3 years ago
11

Which function has an inverse that is also a function? {(–4, 3), (–2, 7), (–1, 0), (4, –3), (11, –7)} {(–4, 6), (–2, 2), (–1, 6)

, (4, 2), (11, 2)} {(–4, 5), (–2, 9), (–1, 8), (4, 8), (11, 4)} {(–4, 4), (–2, –1), (–1, 0), (4, 1), (11, 1)}
Mathematics
2 answers:
KonstantinChe [14]3 years ago
5 0

Answer:  The correct option is

(A) {(–4, 3), (–2, 7), (–1, 0), (4, –3), (11, –7)}.

Step-by-step explanation:  We are given to select the function having an inverse that is also a function.

We know that

a Relation as a set of ordered pairs is a function if each x co-ordinate does not correspond to two different y co-ordinates.

Option (A) : F = {(–4, 3), (–2, 7), (–1, 0), (4, –3), (11, –7)}.

Here, the inverse of F will be

R = {(3, -4), (-2, 7), (0, -1), (-3, 4), (-7, 11)}.

Since no x co-ordinate corresponds to more than one y co-ordinate, so this inverse will be a function.

Option (A) is CORRECT.

Option (B) : F = {(–4, 6), (–2, 2), (–1, 6), (4, 2), (11, 2)}.

Here, the inverse of F will be

R = {(6, -4), (2, -2), (6, -1), (2, 4), (2, 11)}.

Since the x co-ordinate corresponds to three different y co-ordinates (-2, 4 and 11), so this inverse will NOT be a function.

Option (B) is not correct

Option (C) : F =  {(–4, 5), (–2, 9), (–1, 8), (4, 8), (11, 4)}.

Here, the inverse of F will be

R = {(5, -4), (9, -2), (8, -1), (8, 4), (4, 11)}.

Since the x co-ordinate 8 corresponds to two different y co-ordinates (-1 and 4), so this inverse will NOT be a function.

Option (C) is not correct.

Option (D) : F =  {(–4, 4), (–2, –1), (–1, 0), (4, 1), (11, 1)}.

Here, the inverse of F will be

R = {(4, -4), (-1, -2), (0, -1), (1, 4), (1, 11)}.

Since the x co-ordinate 1 corresponds to two different y co-ordinates (4 and 11), so this inverse will NOT be a function.

Option (D) is not correct.

Thus, (A) is the correct option.

DaniilM [7]3 years ago
4 0
(-4,4) because its at the same hold for the other functions
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The function which is same as the function y = 3cos(2(x +π/2)) -2 is: Option A: y= 3sin(2(x + π/4)) - 2

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Here, the given function is:

y= 3\cos(2(x + \pi/2)) - 2

The options are:

  1. y= 3\sin(2(x + \pi/4)) - 2
  2. y= -3\sin(2(x + \pi/4)) - 2
  3. y= 3\cos(2(x + \pi/4)) - 2
  4. y= -3\cos(2(x + \pi/2)) - 2

Checking all the options one by one:

  • Option 1: y= 3\sin(2(x + \pi/4)) - 2

y= 3\sin(2(x + \pi/4)) - 2\\y= 3\sin (2x + \pi/2) -2\\y = -3\cos(2x) -2\\y = 3\cos(2x + \pi) -2\\y = 3\cos(2(x+ \pi/2)) -2

(the last second step was the use of the fact that cos flips its sign after pi radian increment in its input)
Thus, this option is same as the given function.

  • Option 2: y= -3\sin(2(x + \pi/4)) - 2

This option if would be true, then from option 1 and this option, we'd get:
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which isn't true for all values of x.

Thus, this option is not same as the given function.

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The given function is y= 3\cos(2(x + \pi/2)) - 2 = 3\cos(2x + \pi) -2 = -3\cos(2x) -2

This option's function simplifies as:

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  • Option 4: y= -3\cos(2(x + \pi/2)) - 2

The given function simplifies to:y= 3\cos(2(x + \pi/2)) - 2 = 3\cos(2x + \pi) -2 = -3\cos(2x) -2

The given option simplifies to:

y= -3\cos(2(x + \pi/2)) - 2 = -3\cos(2x + \pi ) -2\\y = 3\cos(2x) -2

Thus, this function is not same as the given function.

Thus, the function which is same as the function y = 3cos(2(x +π/2)) -2 is: Option A: y= 3sin(2(x + π/4)) - 2

Learn more about sine to cosine conversion here:

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