Let f(x) = x^3 + ax^2 + bx + c, where a, b, and c are real numbers. If f(-1) = -10 and f(1) = 2, and f'(x) = 0 has real roots, which of the following intervals must contain at least one real root of f(x) = 0?
Correct: C
Given f(-1) = -10 and f(1) = 2, we have -1 + a - b + c = -10 and 1 + a + b + c = 2. Adding the equations, we get 2a + 2c = -8, or a + c = -4. Subtracting, we get 2 + 2b = 12, so b = 5. Thus, f(x) = x^3 + ax^2 + 5x + (-4 - a). Now, f'(x) = 3x^2 + 2ax + 5. Since f'(x) = 0 has real roots, the discriminant must be non-negative: (2a)^2 - 4(3)(5) >= 0, so 4a^2 >= 60, or a^2 >= 15. Thus, a >= sqrt(15) or a <= -sqrt(15). Since f(-1) < 0 and f(1) > 0, there must be a root in (-1, 1). Let's try to narrow down the root by considering the intermediate value theorem. Using a + c = -4, c = -4 - a. f(x) = x^3 + ax^2 + 5x - 4 - a. f(0) = -4 - a. If a > -4, f(0) < 0 and since f(1) > 0 there is a root in (0, 1). If a < -4, f(0) > 0 and since f(-1) < 0 there is a root in (-1, 0). We know a >= sqrt(15) or a <= -sqrt(15). sqrt(15) > 3, so a > 3 or a < -3. Because both (-1,0) and (0,1) are in range (-2, -1), (-1, 0), (0, 1), this is wrong. When we try x=2, f(2) = 8 + 4a + 10 - 4 - a = 14 + 3a. if f(2)>0 and f(1)>0, which it could be when a= -3, so it's not contained (1,2). When we try x=-2, f(-2) = -8 + 4a - 10 - 4 - a = -22 + 3a. if f(-2)<0, and f(-1) < 0, which it could be when a = 7, so it's not contained (-3, -2). so when a=-5, f(0) = -4 -a = 1. f(-1) = -10, so contained in (-1, 0).