Sure, let's find the intervals of concavity for each case. a) f(x) = x^4 - 6x^2 Step 1: Find the second derivative of the function. The first derivative of f(x) is f'(x) = 4x^3 - 12x. The second derivative is f''(x) = 12x^2 - 12. Step 2: Set the second derivative equal to zero and solve for x. This gives us 12x^2 - 12 = 0, which simplifies to x^2 = 1. Therefore, x = -1, 1. Step 3: Test the intervals. We have three intervals to test: (-∞, -1), (-1, 1), and (1, ∞). Choose a test point in each interval and substitute it into the second derivative. If f''(x) 0, the function is concave up on that interval. If f''(x) 0. So, the function is concave up on (-∞, -1). For (-1, 1), let's choose x = 0. Substituting into the second derivative gives 12(0)^2 - 12 = -12 0. So, the function is concave up on (1, ∞). b) f(x) = (x^2 - 1)^3 Step 1: Find the second derivative of the function. The first derivative of f(x) is f'(x) = 3(x^2 - 1)^2 * 2x = 6x(x^2 - 1)^2. The second derivative is f''(x) = 12x(x^2 - 1) + 12x^2 = 12x^3 - 12x. Step 2: Set the second derivative equal to zero and solve for x. This gives us 12x^3 - 12x = 0, which simplifies to x^3 = x. Therefore, x = -1, 0, 1. Step 3: Test the intervals. We have four intervals to test: (-∞, -1), (-1, 0), (0, 1), and (1, ∞). Choose a test point in each interval and substitute it into the second derivative. If f''(x) 0, the function is concave up on that interval. If f''(x)

Question

Sure, let's find the intervals of concavity for each case.

a) f(x) = x^4 - 6x^2

Step 1: Find the second derivative of the function. The first derivative of f(x) is f'(x) = 4x^3 - 12x. The second derivative is f''(x) = 12x^2 - 12.

Step 2: Set the second derivative equal to zero and solve for x. This gives us 12x^2 - 12 = 0, which simplifies to x^2 = 1. Therefore, x = -1, 1.

Step 3: Test the intervals. We have three intervals to test: (-∞, -1), (-1, 1), and (1, ∞). Choose a test point in each interval and substitute it into the second derivative. If f''(x) > 0, the function is concave up on that interval. If f''(x) < 0, the function is concave down.

For (-∞, -1), let's choose x = -2. Substituting into the second derivative gives 12(-2)^2 - 12 = 36 > 0. So, the function is concave up on (-∞, -1).

For (-1, 1), let's choose x = 0. Substituting into the second derivative gives 12(0)^2 - 12 = -12 < 0. So, the function is concave down on (-1, 1).

For (1, ∞), let's choose x = 2. Substituting into the second derivative gives 12(2)^2 - 12 = 36 > 0. So, the function is concave up on (1, ∞).

b) f(x) = (x^2 - 1)^3

Step 1: Find the second derivative of the function. The first derivative of f(x) is f'(x) = 3(x^2 - 1)^2 * 2x = 6x(x^2 - 1)^2. The second derivative is f''(x) = 12x(x^2 - 1) + 12x^2 = 12x^3 - 12x.

Step 2: Set the second derivative equal to zero and solve for x. This gives us 12x^3 - 12x = 0, which simplifies to x^3 = x. Therefore, x = -1, 0, 1.

Step 3: Test the intervals. We have four intervals to test: (-∞, -1), (-1, 0), (0, 1), and (1, ∞). Choose a test point in each interval and substitute it into the second derivative. If f''(x) > 0, the function is concave up on that interval. If f''(x) < 0, the function is concave down.

For (-∞, -1), let's choose x = -2. Substituting into the second derivative gives 12(-2)^3 - 12(-2) = -56 < 0. So, the function is concave down on (-∞, -1).

For (-1, 0), let's choose x = -0.5. Substituting into the second derivative gives 12(-0.5)^3 - 12(-0.5) = 5 > 0. So, the function is concave up on (-1, 0).

For (0, 1), let's choose x = 0.5. Substituting into the second derivative gives 12(0.5)^3 - 12(0.5) = -5 < 0. So, the function is concave down on (0, 1).

For (1, ∞), let's choose x = 2. Substituting into the second derivative gives 12(2)^3 - 12(2) = 56 > 0. So, the function is concave up on (1, ∞).

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