Consider the function g(x)=1/3x+2
a) find and simplify a rule for the inverse. name the inverse g –1. show your work.
b) graph functions g and g –1 on the same coordinate plane. explain how the graphs verify that the rule for g –1 is correct. then explain how the graph verifies that g –1 is a function.
c) use function composition to show that the rule for g –1 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.

 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.

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.

, and

for

step-by-step explanation:

we can analyze the pattern at which the terms increase. let us denote , , , so that we can call

1 the 1st term, 3 the 2nd term, 6 the 3rd term,etc.

note the following:

from the 1st term (1) to the 2nd term (3), we go up by 2,from the 2nd term (3) to the 3rd term (6), we go up by 3from the 3rd term (6) to the 4th term (10), we go up by 4, andfrom the 4th term (10) to the 5th term (15), we go up by 5.

the pattern: to get the next term value, we just add the next term number (e.g. 1st, 2nd, to the previous term value (term values are the given numbers in the question, 1, 3, 6, 10, 15,

from , to get to , we add

from , to get to , we add

etc.

we need to form a recursive formula, a formula that has the next term as a function of the previous term. so a possible recursive formula can be

, and for