Here,
f(x) = |x| + x can be written as
And g(x) = |x| - x, can be written as
Therefore, gof is defined as
For x ≥ 0, gof(x) = g(f(x))
⇒ gof (x) = g(2x) = 0
and for x ≥ 0,gof(x) = g(f(x)) = g(0) =0
Hence, gof (x) = 0 ∀ x ∈ R.
Again,
fog is defined as
For x ≥ 0, fog(x) = f(g(x)) = f(0) = 0
and for x < 0,
fog(x) = f(g(x)) = f(- 2x)
= 2(- 2x) = - 4x
Hence,
2nd part
fog(5) = 0 [∵ 5 ≥ 0]
fog(- 3) = - 4 x (- 3) = 12 [∵ 3 < 0]
gof(-2) = 0