Answer:
For f(x) = x^2 - 2x + 1, the inverse is g(x) = √x + 1
For f(x) = 2 - 5*x the inverse is g(x) = (2 - x)/5
For f(x) = 5*(x - 2) the inverse is g(x) = x*(1/5) + 2
for f(x) = (x - 2)^2 - 1 the inverse is g(x) = √(x + 1) + 2
Explanation:
If we have two functions f(x) and g(x) such that these functions are inverse, then we have:
f( g(x)) = x
and
g( f(x)) = x
Let's start with the function:
f(x) = x^2 - 2x + 1
Notice that we can rewrite this as:
f(x) = x^2 - 2*x + 1 = (x - 1)^2
Because this is a quadratic function, the inverse function needs to have a square root, then the inverse function can be the third or fourth one.
Let's try with the last one:
g(x) = √x + 1
Then:
f( g(x)) = ( √x + 1 - 1)^2 = (√x )^2 = x
g( f(x)) = √(x - 1)^2 + 1 = (x - 1) + 1 = x
Then for the function:
f(x) = x^2 - 2x + 1, the inverse is: g(x) = √x + 1
Now let's find the inverse for the second function:
f(x) = 2 - 5*x
The inverse of this function will be also a linear function, and the slope of the inverse will be the inverse of this slope, then the inverse may be the first or second options.
Let's try with:
g(x) = (1/5)*x - 2
Then:
f( g(x) ) = 2 - 5*((1/5)*x - 2) = 2 - x + 10
Then this g(x) is not the inverse of the function.
Let's try with g(x) = (2 - x)/5
Then:
f( g(x)) = 2 - 5*(2 - x)/5 = 2 - 2 + x = x
g( f(x)) = (2 - 2 - 5*x)/5 = 5*x/5 = x
For the function f(x) = 5*(x - 2)
The inverse should be the remaining linear function:
g(x) = x*(1/5) + 2
Then:
f( g(x)) = 5*(x*(1/5) + 2 - 2) = 5*x*(1/5) = x
g( f(x)) = (5*(x - 2) )*(1/5) + 2 = (x - 2) +2 = x
For the last function, f(x) = (x - 2)^2 - 1
The inverse function will be:
g(x) = √(x + 1) + 2
The compositions are:
f( g(x)) = (√(x + 1) + 2 - 2)^2 - 1 = (√(x + 1))^2 - 1 = (x + 1) - 1 = x
g(f(x)) = √((x - 2)^2 - 1 + 1) + 2 = √((x - 2)^2 + 2 = x