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Given the ordered pairs for f(x) are (3, 7), (4, -1), and (-2, 7), find the ordered pairs for f^-1(x)

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2 points
(3, 7), (4, -1), and (-2, 7)
(7, 3), (-1, 4), and (7, -2)
(-3, -7), (-4, 1), and (2, -7)
If a point is in quadrant II, where will its' inverse be located?
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2 points
I
II
III
IV
Find the inverse of the following:
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2 points
Captionless Image
y = 3(x + 5)
y = 3(x - 5)
y = 1/3x + 5
Is the following set of ordered pairs a function?
(1, 4), (2, 3), (2, 7), and (5, 4)
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2 points
Yes
No

1 answer

  1. Finding \( f^{-1}(x) \) from the ordered pairs:

The inverse function \( f^{-1}(x) \) consists of swapping the x and y values of the original function \( f(x) \). Given the ordered pairs for \( f(x) \):

  • (3, 7) becomes (7, 3)
  • (4, -1) becomes (-1, 4)
  • (-2, 7) becomes (7, -2)

Therefore, the ordered pairs for \( f^{-1}(x) \) are: (7, 3), (-1, 4), and (7, -2).

  1. If a point is in quadrant II, where will its inverse be located?

A point in quadrant II has coordinates \((-x, y)\) where \(x < 0\) and \(y > 0\). Its inverse will have the coordinates \((y, -x)\), which places it in quadrant I because \(y > 0\) and \(-x > 0\) (since \(x\) is negative).

  1. Finding the inverse of the equation:

For the equation \( y = 3(x + 5) \):

  • First, swap \( x \) and \( y \): \[ x = 3(y + 5) \]
  • Then solve for \( y \): \[ y + 5 = \frac{x}{3} \] \[ y = \frac{x}{3} - 5 \]

So, the inverse is: y = \(\frac{1}{3}x - 5\).

(Note: The given options did not include this; please verify that there was no option here for this exact answer.)

  1. Is the following set of ordered pairs a function?

The set of ordered pairs \((1, 4), (2, 3), (2, 7), (5, 4)\) has two different outputs (3 and 7) for the same input (2).

Therefore, this set of ordered pairs is not a function.

In summary:

  1. \( f^{-1}(x) \): (7, 3), (-1, 4), (7, -2)
  2. Quadrant II inverse location: I
  3. Inverse of the equation \( y = 3(x + 5) \): y = \(\frac{1}{3}x - 5\)
  4. Is the set a function? No