Let us try first
derivative test Note that 0 is a critical point of f Now to
the left of 0, f (x) = 3 – x and so f ′(x) = – 1 < 0 Also to
Rationalised 2023-24
APPLICATION OF DERIVATIVES
167
the right of 0, f (x) = 3 + x and so  f ′(x) = 1 > 0