C3 The given example in the image I uploaded should be the format of the solution:

activity 1 LEARNING ACTIVITY Find the value or values of x for which the given function has a maximum or a minimum value. module. 1. y = 8x3 - 9x2 + 1 2. y = x3 - 4x2 + 4x 3. y = 4x-1 + x 4. y = x3 - 3x2 + 3x 5. y = x'(x - 1)2 Second Derivative Test 1. The function y = f(x) is a maximum at x = a if f'(a) = 0 and f"(a) 0. Note that if f"(a) = 0 or if f"(a) does not exist, then SDT fails and under this particular situation, we may use FDT. Example1. Find the value of x for which the function y = x3 - 6x2 + 9x - 3 is a maximum or a minimum. Solution: y = x3 - 6x2 + 9x - 3 y' = 3x2 - 12x + 9 = 3(x2 - 4x + 3) = 3(x - 1)(x -3) y" = 6x - 12 Setting y' = 0, we get x = 1 and x = 3. Then when x = 1, y" 0 Therefore, using SDT, the function is at maximum at x = 1 and a minimum at x = 3. Example 2. Find the value of x for which the function y = (2x - 1) is a maximum or a minimum. Solution: y = (2x - 1)2 y' = 2(2x - 1) . 2 = 4(2x - 1) = 8x - 4 y" = 8 Setting y' = 0, we get x = = when x = = y' = 0 and y" > 0 Therefore, using SDT, the function is at minimum at x = = LEARNING ACTIVITY Find the value or values of x for which the given function has a maximum or a minimum value. module. 1. y = 2x3 - 3x2 - 36x + 25 2. y = 3x4 - 4x3+ 1 3. y = x4 - 4x3 + 4x2 4. y = 3x5 - 15x4 + 20x3 + 3 activity 2