Consider the function g(x) shown below over a domain of [1,∞).

g(x) = 2(x – 1)2 – 3

Part A
Determine g–1(x).


Part B
Identify the domain and range of g(x) and g–1(x). Represent the domains and ranges in inequality, interval, and set notation. Describe the relationship between the domains and ranges of g(x) and g–1(x).


Part C
Graph g(x) and g–1(x). Describe the relationship between the graphs of g(x) and g–1(x).

The inverse of a function may or may not be a function

The function g(x) is given as:

[tex]g(x) = 2(x - 1)^2 - 3[/tex]

(a) Inverse function g^-1(x)

We have:

[tex]g(x) = 2(x - 1)^2 - 3[/tex]

Replace g(x) with y

[tex]y = 2(x - 1)^2 - 3[/tex]

Swap the positions of x and y

[tex]x = 2(y - 1)^2 - 3[/tex]

Add 3 to both sides

[tex]x + 3 = 2(y - 1)^2[/tex]

Divide both sides by 2

[tex]\frac{x + 3}2 = (y - 1)^2[/tex]

Take square roots of both sides

[tex]\sqrt{\frac{x + 3}2} = y - 1[/tex]

Add 1 to both sides

[tex]1 + \sqrt{\frac{x + 3}2} = y[/tex]

Rewrite as:

[tex]y = 1 + \sqrt{\frac{x + 3}2}[/tex]

Express as inverse function of g

[tex]g^{-1}(x) = 1 + \sqrt{\frac{x + 3}2}[/tex]

Hence, the inverse function of g is [tex]g^{-1}(x) = 1 + \sqrt{\frac{x + 3}2}[/tex]

(b) The domain, and the range of g(x) and g^-1(x)

Function g(x)

The function g(x) is a quadratic function.

So, the domain is:

The range is:

Function g^-1(x)

The function g^-1(x) is a square root function.

So, the domain is:

The range is:

The domain of g(x) is the range of g^-1(x), while the range of g(x) is the domain of g^-1(x).

(c) The graph

See attachment for the graph

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