It is known that a is a constant, and a > 0, vector M = (√ x, - 1), vector n = (1, ln (x + a)), Finding the maximum value of the function f (x) = m · n in the interval (0,1]

It is known that a is a constant, and a > 0, vector M = (√ x, - 1), vector n = (1, ln (x + a)), Finding the maximum value of the function f (x) = m · n in the interval (0,1]

F (x) = M &; n = √ x-ln (x + a) f '(x) = 1 / (2 * √ (x)) - 1 / (x + a) = (x + A-2 * √ (x)) / (2 * √ (x) * (x + a)) according to the condition x > 0, a > 0, so the denominator is greater than 0, only need to observe the molecule. Let √ (x) = t (T > 0), so x = T ^ 2. X + A-2 * √ (x) is equivalent to T ^ 2-2t + A, and y = T ^ 2-2t + A. m discriminant △