Function f (x) = 2lnx - X & # 178; - KX (K ∈ R), if function f (x) has two zeros m, n (0 ＜ m ＜ n), and 2x0 = m + n. question: can the tangent of function f (x) at the point (x0, f (x0)) be parallel to the X axis? If so, find the tangent equation; if not, explain the reason

Function f (x) = 2lnx - X & # 178; - KX (K ∈ R), if function f (x) has two zeros m, n (0 ＜ m ＜ n), and 2x0 = m + n. question: can the tangent of function f (x) at the point (x0, f (x0)) be parallel to the X axis? If so, find the tangent equation; if not, explain the reason

Suppose: the tangent of the function f (x) at the point (x0, f (x0)) is parallel to the x-axis. According to the meaning of the question, we get: 2lnm-m & # 178; - km = 0, ① 2lnn-n & # 178; - kn = 0, ② m + n = 2x0, ③ 2 / x0-2x0-k = 0, ④ ① - ② get: ln (M / N) - (M + n) (m-n) = K (m-n) ‖ k = [2ln (M / N)] / (m-n) - 2x0. From the variation of ④, we get