Let f(x) = 1 4 (4x) + 1 2 (2x) - 1 . The number of points in the interval x [0, 5 ] where the tangent to the curve y = f(x) is horizontal, is ______.

Step-by-step solution

For the tangent to be horizontal, the derivative of the function must be zero. That is, f'(x) = 0 . Given f(x) = 1 4 (4x) + 1 2 (2x) - 1 , we differentiate with respect to x : f'(x) = 1 4 (-4 (4x)) + 1 2 (-2 (2x)) f'(x) = - (4x) - (2x) Setting f'(x) = 0 gives: (4x) + (2x) = 0 Using the double angle identity (4x) = 2 (2x) (2x) , we get: 2 (2x) (2x) + (2x) = 0 (2x)(2 (2x) + 1) = 0 This gives two cases: Case 1: (2x) = 0 Case 2: (2x) = - 1 2 The given interval for x is [0, 5 ] , which means 2x [0, 10 ] . For Case 1: (2