The function y = 2x3 + ax2 + bx has critical points at x=2 and x=1. Find the values of a and b, the values of x for which y is minimum, and the minimum value of y.

Hey B🧡i🧡b🧡i💛, good to see you again. Here's the solution to question 1.a. a) What do you mean by the minimum point of a function? The function y = 2x^3 + ax^2 + bx where a and b are constants, has critical points at x=2 and x=1. Find: (i) The values of a and b. (ii) The values of x for which y is minimum. (iii) The minimum value of y. Step 1: Define the minimum point of a function. A minimum point of a function is a point where the function's value is less than or equal to its values at nearby points. At a minimum point, the first derivative of the function is zero, and the second derivative is positive. Step 2: Find the first derivative of the function. The given function is y = 2x^3 + ax^2 + bx. To find the critical points, we need to find the first derivative (dy)/(dx) and set it to zero. (dy)/(dx) = (d)/(dx)(2x^3 + ax^2 + bx) = 6x^2 + 2ax + b Step 3: Use the given critical points to form equations for a and b. We are given that the critical points are at x=2 and x=1. This means (dy)/(dx) = 0 at these points. For x=2: 6(2)^2 + 2a(2) + b = 0 6(4) + 4a + b = 0 24 + 4a + b = 0 (*1) For x=1: 6(1)^2 + 2a(1) + b = 0 6 + 2a + b = 0 (*2) Step 4: Solve the system of equations for a and b. Subtract equation (2) from equation (1): (24 + 4a + b) - (6 + 2a + b) = 0 - 0 24 - 6 + 4a - 2a + b - b = 0 18 + 2a = 0 2a = -18 a = -9 Substitute a = -9 into equation (*2): 6 + 2(-9) + b = 0 6 - 18 + b = 0 -12 + b = 0 b = 12 a) (i) The values of a and b: a = -9, b = 12 Step 5: Find the second derivative to determine the nature of the critical points. The first derivative is (dy)/(dx) = 6x^2 + 2ax + b. Substitute the values of a and b: (dy)/(dx) = 6x^2 + 2(-9)x + 12 = 6x^2 - 18x + 12 Now, find the second derivative (d^2y)/(dx^2): (d^2y)/(dx^2) = (d)/(dx)(6x^2 - 18x + 12) = 12x - 18 Step 6: Evaluate the second derivative at each critical point. For x=2: (d^2y)/(dx^2) |_x=2 = 12(2) - 18 = 24 - 18 = 6 Since (d^2y)/(dx^2) > 0 at x=2, this point is a minimum. For x=1: (d^2y)/(dx^2) |_x=1 = 12(1) - 18 = 12 - 18 = -6 Since (d^2y)/(dx^2) < 0 at x=1, this point is a maximum. a) (ii) The values of x for which y is minimum: x = 2 Step 7: Calculate the minimum value of y. Substitute x=2, a=-9, and b=12 into the original function y = 2x^3 + ax^2 + bx. y_min = 2(2)^3 + (-9)(2)^2 + (12)(2) y_min = 2(8) - 9(4) + 24 y_min = 16 - 36 + 24 y_min = -20 + 24 y_min = 4 a) (iii) The minimum value of y: y_min = 4 Send me the next one 📸