The functions f(x) = √(x+2) and g(x) =(x^2 - 3)/2 are inverses of each other, as given by: Option C: Functions f and g are inverses because f(g(x)) = g(f(x)) (for any real number x)
Suppose that the given function is
[tex]f:X\rightarrow Y[/tex]
Then, if function 'f' is one-to-one and onto function (a needed condition for inverses to exist), then, the inverse of the considered function is
[tex]f^{-1}: Y \rightarrow X[/tex]
such that:
[tex]\forall \: x \in X : f(x) \in Y, \exists \: y \in Y : f^{-1}(y) \in X[/tex]
(and vice versa).
It simply means, inverse of 'f' is undo operator, that takes back the effect of 'f'
For this case, we're specified that:
If f and g are inverses of each other, then:
[tex]f(g(x)) = f(f^{-1}(x)) = x \: \forall \: x \in \text{Domain of g(x) }\\\\g(f(x)) = g(g^{-1}(x)) = x \: \forall \: x \in \text{Domain of f(x) }[/tex]
(if these two conditions are met, then it can be proved that f and g are inverse of each other)
[tex]f(g(x)) = \sqrt{2g(x) + 2} = \sqrt{ 2\left(\dfrac{x^2 - 2}{2}\right) + 2} = \sqrt{x^2} = x[/tex]
[tex]g(f(x)) = \dfrac{(f(x))^2 -2}{2} = \dfrac{(\sqrt{2x-2})^2 -2}{2} = \dfrac{2x}{2} = x[/tex]
In both evaluation, all the steps are valid for any real number 'x'.
[tex]f(g(x)) = x = g(f(x)) \: \forall \: x \in \mathbb R[/tex]
(which is superset of both domain and range of both functions, assuming those functions are having their domain and range as subsets of real number set).
Thus, the functions f(x) = √(x+2) and g(x) =(x^2 - 3)/2 are inverses of each other, as given by: Option C: Functions f and g are inverses because f(g(x)) = g(f(x)) (for any real number x)
Learn more about inverse function here:
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