This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Here's the solution for the second problem. Problem 2: Determine the critical points, their nature, and sketch the graph for y = 6x^3 + x^2 - 4x + 1. Step 1: Find the first derivative. To find the critical points, we first calculate the first derivative of the function y with respect to x. y' = (d)/(dx)(6x^3 + x^2 - 4x + 1) y' = 6(3x^2) + 2x - 4 y' = 18x^2 + 2x - 4 Step 2: Determine the critical points. Set the first derivative equal to zero to find the x-coordinates of the critical points. 18x^2 + 2x - 4 = 0 Divide the entire equation by 2 to simplify: 9x^2 + x - 2 = 0 This is a quadratic equation. We can use the quadratic formula x = -b ± sqrt(b^2 - 4ac)2a where a=9, b=1, c=-2. x = -1 ± sqrt(1^2 - 4(9)(-2))2(9) x = -1 ± sqrt(1 + 72)18 x = -1 ± sqrt(73)18 This gives two critical x-values: x_1 = -1 + sqrt(73)18 ≈ (-1 + 8.544)/(18) ≈ (7.544)/(18) ≈ 0.419 x_2 = -1 - sqrt(73)18 ≈ (-1 - 8.544)/(18) ≈ (-9.544)/(18) ≈ -0.530 Now, substitute these x-values back into the original function y = 6x^3 + x^2 - 4x + 1 to find the corresponding y-values. For x_1 = -1 + sqrt(73)18 ≈ 0.419: y_1 = 6(-1 + sqrt(73)18)^3 + (-1 + sqrt(73)18)^2 - 4(-1 + sqrt(73)18) + 1 y_1 ≈ 6(0.419)^3 + (0.419)^2 - 4(0.419) + 1 y_1 ≈ 6(0.0736) + 0.1756 - 1.676 + 1 y_1 ≈ 0.4416 + 0.1756 - 1.676 + 1 ≈ -0.0588 The first critical point is approximately (0.419, -0.059). For x_2 = -1 - sqrt(73)18 ≈ -0.530: y_2 = 6(-1 - sqrt(73)18)^3 + (-1 - sqrt(73)18)^2 - 4(-1 - sqrt(73)18) + 1 y_2 ≈ 6(-0.530)^3 + (-0.530)^2 - 4(-0.530) + 1 y_2 ≈ 6(-0.1488) + 0.2809 + 2.12 + 1 y_2 ≈ -0.8928 + 0.2809 + 2.12 + 1 ≈ 2.5081 The second critical point is approximately (-0.530, 2.508). The critical points are (-1 + sqrt(73)18, -0.059) and (-1 - sqrt(73)18, 2.508). Step 3: Determine the nature of the critical points. To determine if these points are local maxima or minima, we use the second derivative test. First, find the second derivative: y'' = (d)/(dx)(18x^2 + 2x - 4) y'' = 36x + 2 Now, evaluate y'' at each critical x-value: For x_1 = -1 + sqrt(73)18 ≈ 0.419: y''(-1 + sqrt(73)18) = 36(-1 + sqrt(73)18) + 2 y''(-1 + sqrt(73)18) = 2(-1 + sqrt(73)) + 2 y''(-1 + sqrt(73)18) = -2 + 2sqrt(73) + 2 = 2sqrt(73) Since 2sqrt(73) > 0, the critical point (-1 + sqrt(73)18, -0.059) is a local minimum. For x_2 = -1 - sqrt(73)18 ≈ -0.530: y''(-1 - sqrt(73)18) = 36(-1 - sqrt(73)18) + 2 y''(-1 - sqrt(73)18) = 2(-1 - sqrt(73)) + 2 y''(-1 - sqrt(73)18) = -2 - 2sqrt(73) + 2 = -2sqrt(73) Since -2sqrt(73) < 0, the critical point (-1 - sqrt(73)18, 2.508) is a local maximum. Step 4: Sketch the graph. The function is a cubic polynomial with a positive leading coefficient (6x^3). This means the graph will generally rise from left to right. • As x -, y -. • As x , y . The graph will increase until it reaches the local maximum at approximately (-0.530, 2.508), then decrease until it reaches the local minimum at approximately (0.419, -0.059), and finally increase indefinitely. Drop the next question.