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:
- Inequality: [tex]-\infty < x < \infty[/tex]
- Interval: [tex](-\infty ,\infty)[/tex]
- Set: {R}
The range is:
- Inequality: [tex]f(x) \ge -3[/tex]
- Interval: [tex][-3 ,\infty)[/tex]
- Set: [tex]\{y|y\ge -3\}[/tex]
Function g^-1(x)
The function g^-1(x) is a square root function.
So, the domain is:
- Inequality: [tex]x \ge -3[/tex]
- Interval: [tex][-3 ,\infty)[/tex]
- Set: [tex]\{x|x\ge -3\}[/tex]
The range is:
- Inequality: [tex]-\infty < y < \infty[/tex]
- Interval: [tex](-\infty ,\infty)[/tex]
- Set: {R}
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
Read more about inverse functions at:
https://brainly.com/question/14391067
