If f, g: R → R be two functions defined as f(x) = |x| + x and
g(x) = |x|–x for all x ∈ R. Then, find fog and gof. Hence, find fog (–3),
fog (5) and gof(–2).
Domain of f(x) and g(x) is R.
Range of f(x) = [0, ∞) and range of g(x) = [0, ∞)
As, range of f ⊂ Domain of g and range of g ⊂ Domain of f
So, gof and fog exists
Now,
fog(x) = f(g(x)) = f(|x|–x)
⇒ fog(x) = ||x|–x| + |x|–x
As, range of g(x) ≥ 0 so, ||x|–x| = |x|–x
So, fog(x) = ||x|–x| + |x|–x = |x|–x + |x|–x
⇒ fog(x) = 2(|x|–x)
Also,
gof(x) = g(f(x)) = g(|x| + x) = ||x| + x| – (|x| + x)
As, range of f(x) ≥ 0 so, ||x| + x| = |x| + x
So, gof(x) = ||x| + x| – (|x| + x) = |x| + x – (|x| + x) = 0
Thus, gof(x) = 0
Now, fog(– 3) = 2(| – 3|–(– 3)) = 2(3 + 3) = 6,
fog(5) = 2(|5| – 5) = 0, gof(– 2) =0