f(x) = x² + 4x + 1 has a A) relative maximum at x = -2 B) relative minimum at x = -2 C) relative maximum at x = -4 D) relative minimum at x = -4 E) relative maximum at x = -1

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To find critical points, we take the derivative of the function and set it to 0. f'(x) = 3² x^2 + 4 Setting f'(x) = 0, we get 3² x^2 + 4 = 0 x^2 = -4/(3²) Since the discriminant is negative, this means there are no real roots and hence no critical points. Therefore, the function has no relative maximum or minimum. However, we can observe that the coefficient of x^2 is positive, indicating that the function opens upwards and hence has a minimum. Since the function is increasing for all x, this minimum occurs at the leftmost point on the domain, which is x = -2. Therefore, the function has a relative minimum at x = -2.

F(x) = X2 + 4x + 1 Has a

F(x) = X² + 4x + 1 Has a