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.
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.
