Answer: C) x = 3, (x = 2 is an extraneous solution)
<u>Step-by-step explanation:</u>
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= 
- = 
Restrictions: <em>Denominator cannot equal zero</em> (x - 1 ≠ 0) and (x - 2 ≠ 0), so x ≠ 1 and x ≠ 2
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= 
x(x - 2) - 1(x - 1) = 2x - 5
x² - 2x - x + 1 = 2x - 5
x² - 3x + 1 = 2x - 5
<u> -2x + 5</u> <u>-2x + 5 </u>
x² - 5x + 6 = 0
(x - 2)(x - 3) = 0
x - 2 = 0 x - 3 = 0
x = 2 x = 3
NOTE: x = 2 is an extraneous solution because it is one of the restricted values.
Answer:
D
Step-by-step explanation:
0.36|6=0.366
Answer:
X
=
27
Step-by-step explanation:
Given:
g(x) = (1/3)x + 2
Part (a)
To find the inverse:
Set y = g(x) = (1/3)x + 2
Swap x and y.
x = (1/3)y + 2.
Solve for y.
(1/3)y = x - 2
y = 3(x - 2).
Set g⁻¹(x) to y.
Answer: g⁻¹(x) = 3(x - 2)
Part (b)
Create the table shown below to graph g(x) and g⁻¹(x).
x g(x) g⁻¹(x)
---- --------- ---------
-8 - 2/3 - 30
-6 0 - 24
-4 2/3 - 18
-2 4/3 - 12
0 2 - 6
2 8/3 0
4 4/3 6
6 4 12
8 14/3 18
Note that when x = -6, g(x) = 0, so that (-6, 0) lies on he black liine.
Therefore the inverse function should yield (0, -6) to be correct. This is so, so g⁻¹ is correct.
Both g(x) and g⁻¹(x)satisfy the vertical line test, so both are functions.
Part (c)
Algebraically, we know that g⁻¹(x) is correct if g(g⁻¹(x)) = x
Use function composition to obtain
g(g⁻¹(x)) = (1/3)*(3x - 6) + 2
= x - 2 + 2
= x
Therefore g⁻¹(x) is correct.
Answer:
The answers are A and B.
Step-by-step explanation:
