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If f: R -> R and g: R ->R are two given ...

If `f: R -> R` and `g: R ->R` are two given functions, then prove that `2min {f(x)-g(x),0}=f(x)-g(x)-|g(x)-f(x)|`

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The correct Answer is:
D
`f : R rarr R, g : R rarr R`
f(x) = 2 min f(x) - g(x), 0
Let f(x) - g(x) `gt` 0, then
F(x) = f(x) - g(x) - |f(x) - g(x)| and f(x) - g(x) `lt` 0, then
F(x) = 2[F(x) - g(x)] = [f(x) - g(x)] - |f(x) - g(x)|
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