X^3+x^2-x-1thus determine the stationary points and their nature
Explore the analysis of stationary points in the cubic function f(x) = x^3 + x^2 - x - 1. Learn how to find and interpret these points, including their nature as maximum or minimum points. Gain insights into the derivative and second derivative to determine the behavior of the function around these critical points.
To determine the stationary points and their nature for the function f(x) = x^3 + x^2 - x - 1, we need to find the derivative and solve for values of x where the derivative equals zero.
First, let's find the derivative of f(x): f'(x) = 3x^2 + 2x - 1 Next, we equate the derivative to zero and solve for x: 3x^2 + 2x - 1 = 0 This equation can be solved using various methods, such as factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula in this case: x = (-b ± √(b^2 - 4ac)) / (2a) For our equation, a = 3, b = 2, and c = -1. Plugging in these values: x = (-2 ± √(2^2 - 43(-1))) / (2*3) x = (-2 ± √(4 + 12)) / 6 x = (-2 ± √16) / 6 x = (-2 ± 4) / 6 This gives us two possible values for x: x1 = (-2 + 4) / 6 = 2/6 = 1/3 x2 = (-2 - 4) / 6 = -6/6 = -1 Now, we have two potential stationary points: (1/3, f(1/3)) and (-1, f(-1)). To determine their nature (whether they are maximum, minimum, or inflection points), we can analyze the second derivative or observe the behavior of the function around these points. Taking the second derivative of f(x): f''(x) = 6x + 2 Substituting x = 1/3: f''(1/3) = 6(1/3) + 2 = 2 + 2 = 4 Since the second derivative is positive, f(1/3) is a minimum point. For x = -1: f''(-1) = 6(-1) + 2 = -6 + 2 = -4 Since the second derivative is negative, f(-1) is a maximum point. Therefore, the nature of the stationary points is as follows:
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