If 1 Δ k = 3, then the range of function f (x) = k Δ x is

√k+1+k=3
k+√k-2=0
So k = 1
So f (x) = √ x + X + 1
=(√x+1/2)²+3/4
√x≥0
So x = 0, the minimum is 1
Range [1, + ∞)

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