Mathematics

Mathematics

Problem

Let f(x)=2⌊x⌋+∣x∣f(x)=2\lfloor x\rfloor+|x|. Evaluate: a. (i) f(−2.3)f(-2.3) b. (ii) f(3)f(3)
Intuitive explanation

The key is to recognize that the floor function and absolute value behave differently on different intervals, so you'll need to split the domain into cases. On any interval [n,n+1)[n, n+1) where nn is an integer, ⌊x⌋\lfloor x \rfloor is constant, but ∣x∣|x| changes its form at x=0x=0. By carefully evaluating ff on each relevant piece, you'll see a piecewise pattern emerge that makes the evaluation straightforward.

Intuitive explanation

The key is to recognize that the floor function and absolute value behave differently on different intervals, so you'll need to split the domain into cases. On any interval [n,n+1)[n, n+1) where nn is an integer, ⌊x⌋\lfloor x \rfloor is constant, but ∣x∣|x| changes its form at x=0x=0. By carefully evaluating ff on each relevant piece, you'll see a piecewise pattern emerge that makes the evaluation straightforward.

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