Consider the function f: R--> R where 𝑐1 , 𝑐2 ∈ ℝ If 𝑓 is continuous at 𝑥…

For x ≤ 0, f(x) = 3, so f(0) = 3 and the left-hand limit at 0 is 3. For x > 0, f(x) = c₁eˣ - c₂ ln(1/x). As x → 0+, ln(1/x) → ∞. For the right-hand limit to be finite, c₂ must be 0. Then the right-hand limit becomes c₁e⁰ = c₁. Continuity at x = 0 requires c₁ = 3. Therefore c₁ + c₂ = 3 + 0 = 3.