Mathematics 2011 JAMB Past Questions - Question 38: Find the value of x at the minimum point of the curve y= x³ + x² -x + 1

Correct Answer

D

Explanation

To find the value of x at the minimum point of the curve y = x³ + x² - x + 1, we can take the derivative of the function and set it equal to zero.Let's find the derivative of y with respect to x:To find the value of x at the minimum point of the curve y = x³ + x² - x + 1, we can find the derivative of the function and set it equal to zero.Let's find the derivative of y with respect to x:dy/dx = 3x² + 2x - 1To find the minimum point, we set the derivative equal to zero and solve for x:3x² + 2x - 1 = 0This is a quadratic equation, which can be factored or solved using the quadratic formula. In this case, let's use the quadratic formula:x = (-b ± √(b² - 4ac)) / (2a)Plugging in the values a = 3, b = 2, and c = -1, we have:x = (-2 ± √(2² - 4(3)(-1))) / (2(3))Simplifying further:x = (-2 ± √(4 + 12)) / 6x = (-2 ± √16) / 6x = (-2 ± 4) / 6This gives us two possible values for x:x₁ = (-2 + 4) / 6 = 2/6 = 1/3x₂ = (-2 - 4) / 6 = -6/6 = -1Since we are looking for the minimum point, we take the smaller value of x, which is x = -1.Therefore, the value of x at the minimum point of the curve y = x³ + x² - x + 1 is x = -1.dy/dx = 3x² + 2x - 1To find the minimum point, we set the derivative equal to zero and solve for x:3x² + 2x - 1 = 0This is a quadratic equation, which can be factored or solved using the quadratic formula. In this case, let's use the quadratic formula:x = (-b ± √(b² - 4ac)) / (2a)Plugging in the values a = 3, b = 2, and c = -1, we have:x = (-2 ± √(2² - 4(3)(-1))) / (2(3))Simplifying further:x = (-2 ± √(4 + 12)) / 6x = (-2 ± √16) / 6x = (-2 ± 4) / 6This gives us two possible values for x:x₁ = (-2 + 4) / 6 = 2/6 = 1/3x₂ = (-2 - 4) / 6 = -6/6 = -1Since we are looking for the minimum point, we take the smaller value of x, which is x = -1.Therefore, the value of x at the minimum point of the curve y = x³ + x² - x + 1 is x = -1.