Consider the function g(x)=1/3x+2
2 answers:
Part A.
The given equation is g(x) = 1/3x + 2. First, let g(x) be y so that it would be easy to write in the solution.
y = 1/3x + 2
Next, interchange x and y variables:
x = 1/3y + 2
Solve for y
1/3y = x - 2
y = 3(x-2) = g⁻¹(x) ← This is the inverse function of the original equation
Part B. Graph both equations by assigning arbitrary values of x. You will get their corresponding y values. Plot x against y, then connect the dots. The graph is shown in the attached picture. It is proven that they are inverse functions if when point (a,b) is plotted on g(x), then point (b,a) must also lie on g⁻¹(x). As shown in the picture, P(6,4) in g(x) is the inverse of P(4,6) in g⁻¹(x).
Part C. The proof of inverse functions in functional notation is:
g(g⁻¹(x)) = x and g⁻¹(g(x) = x, for all values of x. Applying this rule,
g(x)=1/3x+2; g⁻¹(x) = 3(x-2)
g(3(x-2) ) = 1/3(3(x-2) )+2 = x - 2 + 2 = x
and
g⁻¹(1/3x+2 ) = 3(1/3x+2-2) = x +6 - 6 = x
Therefore, the equation for g⁻¹(x) is correct.
The given equation is g(x) = 1/3x + 2. First, let g(x) be y so that it would be easy to write in the solution.
y = 1/3x + 2
Next, interchange x and y variables:
x = 1/3y + 2
Solve for y
1/3y = x - 2
y = 3(x-2) = g⁻¹(x) ← This is the inverse function of the original equation
Part B. Graph both equations by assigning arbitrary values of x. You will get their corresponding y values. Plot x against y, then connect the dots. The graph is shown in the attached picture. It is proven that they are inverse functions if when point (a,b) is plotted on g(x), then point (b,a) must also lie on g⁻¹(x). As shown in the picture, P(6,4) in g(x) is the inverse of P(4,6) in g⁻¹(x).
Part C. The proof of inverse functions in functional notation is:
g(g⁻¹(x)) = x and g⁻¹(g(x) = x, for all values of x. Applying this rule,
g(x)=1/3x+2; g⁻¹(x) = 3(x-2)
g(3(x-2) ) = 1/3(3(x-2) )+2 = x - 2 + 2 = x
and
g⁻¹(1/3x+2 ) = 3(1/3x+2-2) = x +6 - 6 = x
Therefore, the equation for g⁻¹(x) is correct.
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.
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