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Problem of finding out fog and gof can also be handled using graphical methods
f(g(x))
Here g(x) becomes the variable that means we would draw the graph of g(x). It is clear that g(x) < 1 ∀ x ε [-1, 1] and 1 < g(x) < 2 ∀ x ε (1, √2]
In this case f(x) becomes the variable and we will draw the graph of f(x). From the graph we observe that -1 < f(x) < 2 ∀ x ε [-2, 1) and f(x) = x + 1.
2 < f(x)) < 3 => x = 1 and f(x) = x + 1.
i.e. g(f(x)) = (x + 1)2, -2 < x < 1
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