Two functions f and g are defined on the set of real numbers, R, by f:x → x^(2) + 2 and g:x → (1)/(x+2). Find the domain of (g.f)^(-1)

Question

Two functions f and g are defined on the set of real numbers, R, by

f:x →
`x^(2)` + 2 and g:x →
`(1)/(x+2)`. Find the domain of
`(g.f)^(-1)`

Answer 1

To find the domain of (g.f)^-1, we need to find the composition of g and f, then find the inverse, and determine the domain of the inverse. The domain is all real numbers such that x is not equal to 0 and x is greater than 1/4.

Step-by-step explanation:

The domain of a function is the set of all possible input values for the function. To find the domain of (g.f)^{-1}, we first need to find the composition of g and f. The composition of two functions f and g is denoted as (g.f)(x) and it means that you first apply f to x and then apply g to the result. So in this case, (g.f)(x) = g(f(x)). Let's calculate it:

(g.f)(x) = g(f(x)) = g(x^2 + 2) = \frac{1}{x^2 + 2 + 2} = \frac{1}{x^2 + 4}

Now, to find the inverse of (g.f), we need to swap the roles of x and y and solve for y. Let's do it:

y = \frac{1}{x^2 + 4} \implies x = \frac{1}{y^2 + 4} \implies y^2 + 4 = \frac{1}{x} \implies y^2 = \frac{1}{x} - 4 \implies y = \sqrt{\frac{1}{x} - 4}

Therefore, the domain of (g.f)^{-1} is the set of all values of x for which \frac{1}{x} - 4 is positive, and also excluding any x values that would make \frac{1}{x} - 4 equal to zero since division by zero is undefined. In other words, the domain is all real numbers such that x \\eq 0 and x > \frac{1}{4}.