Question

Consider the function 𝑓: ℝ→ℝ defined as follows:
𝑓(𝑥) = {𝑐1𝑒𝑥−𝑐2 loge (1
𝑥) , if 𝑥> 0
3 otherwise
where 𝑐1 , 𝑐2 ∈ℝ.
If 𝑓 is continuous at 𝑥 = 0, then 𝑐1 + 𝑐2 = _________. (answer in integer)

Hint:

For continuity, the right-hand limit as x tends to 0 must equal f(0)=3; since log_e x diverges at x=0, the coefficient of the log term must be zero.

Answer 1

Correct Answer: 3

Step 1: For \(x>0\), \(f(x) = c_1 e^x - c_2 \log_e\left(\frac{1}{x}\right)\). Since \(\log_e\left(\frac{1}{x}\right) = -\log_e x\), this simplifies to \(f(x) = c_1 e^x + c_2 \log_e x\) for \(x>0\).

Step 2: Continuity of \(f\) at \(x=0\) requires \(\lim_{x \to 0^+} f(x) = f(0) = 3\).

Step 3: As \(x \to 0^+\), \(\log_e x \to -\infty\). If \(c_2 \neq 0\), the term \(c_2 \log_e x\) blows up to \(+\infty\) or \(-\infty\), so the limit would not be finite. Continuity therefore forces \(c_2 = 0\).

Step 4: With \(c_2 = 0\), \(\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} c_1 e^x = c_1 e^0 = c_1\).

Step 5: Setting this equal to \(f(0)=3\) gives \(c_1 = 3\). Therefore \(c_1 + c_2 = 3 + 0 = 3\).

Final Answer: \[\boxed{c_1 + c_2 = 3}\]