(a) To find the interval(s) on which f is increasing, we need to determine where the derivative of f is positive. Let's find the derivative of f(x) first. f'(x) = 9 cos(x) - 9 sin(x) To find where f'(x) 0, we need to solve the inequality: 9 cos(x) - 9 sin(x) 0 Dividing both sides by 9, we get: cos(x) - sin(x) 0 Now, we can analyze the sign of cos(x) - sin(x) in different intervals. For 0 ≤ x ≤ 2𝜋, we can divide this interval into smaller intervals based on the behavior of cos(x) and sin(x). In the interval 0 ≤ x ≤ 𝜋/4, both cos(x) and sin(x) are positive. Therefore, cos(x) - sin(x) 0 in this interval. In the interval 𝜋/4 ≤ x ≤ 3𝜋/4, cos(x) is positive and sin(x) is negative. Therefore, cos(x) - sin(x) 0 in this interval. In the interval 3𝜋/4 ≤ x ≤ 5𝜋/4, both cos(x) and sin(x) are negative. Therefore, cos(x) - sin(x) 0, we have: 0 ≤ x ≤ 𝜋/4 ∪ 𝜋/4 ≤ x ≤ 3𝜋/4 ∪ 7𝜋/4 ≤ x ≤ 2𝜋 So, the interval(s) on which f is increasing is [0, 𝜋/4] ∪ [𝜋/4, 3𝜋/4] ∪ [7𝜋/4, 2𝜋]. (b) To find the interval(s) on which f is decreasing, we need to determine where the derivative of f is negative. Let's find the derivative of f(x) again. f'(x) = 9 cos(x) - 9 sin(x) To find where f'(x)

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(a) To find the interval(s) on which f is increasing, we need to determine where the derivative of f is positive. Let's find the derivative of f(x) first. f'(x) = 9 cos(x) - 9 sin(x) To find where f'(x) > 0, we need to solve the inequality: 9 cos(x) - 9 sin(x) > 0 Dividing both sides by 9, we get: cos(x) - sin(x) > 0 Now, we can analyze the sign of cos(x) - sin(x) in different intervals. For 0 ≤ x ≤ 2𝜋, we can divide this interval into smaller intervals based on the behavior of cos(x) and sin(x). In the interval 0 ≤ x ≤ 𝜋/4, both cos(x) and sin(x) are positive. Therefore, cos(x) - sin(x) > 0 in this interval. In the interval 𝜋/4 ≤ x ≤ 3𝜋/4, cos(x) is positive and sin(x) is negative. Therefore, cos(x) - sin(x) > 0 in this interval. In the interval 3𝜋/4 ≤ x ≤ 5𝜋/4, both cos(x) and sin(x) are negative. Therefore, cos(x) - sin(x) < 0 in this interval. In the interval 5𝜋/4 ≤ x ≤ 7𝜋/4, cos(x) is negative and sin(x) is positive. Therefore, cos(x) - sin(x) < 0 in this interval. In the interval 7𝜋/4 ≤ x ≤ 2𝜋, both cos(x) and sin(x) are positive. Therefore, cos(x) - sin(x) > 0 in this interval. Combining all the intervals where cos(x) - sin(x) > 0, we have: 0 ≤ x ≤ 𝜋/4 ∪ 𝜋/4 ≤ x ≤ 3𝜋/4 ∪ 7𝜋/4 ≤ x ≤ 2𝜋 So, the interval(s) on which f is increasing is [0, 𝜋/4] ∪ [𝜋/4, 3𝜋/4] ∪ [7𝜋/4, 2𝜋]. (b) To find the interval(s) on which f is decreasing, we need to determine where the derivative of f is negative. Let's find the derivative of f(x) again. f'(x) = 9 cos(x) - 9 sin(x) To find where f'(x) < 0, we need to solve the inequality: 9 cos(x) - 9 sin(x) < 0 Dividing both sides by 9, we get: cos(x) - sin(x) < 0 Now, we can analyze the sign of cos(x) - sin(x) in different intervals. For 0 ≤ x ≤ 2𝜋, we can divide this interval into smaller intervals based on the behavior of cos(x) and sin(x). In the interval 0 ≤ x ≤ 𝜋/4, both cos(x) and sin(x) are positive. Therefore, cos(x) - sin(x) > 0 in this interval. In the interval 𝜋/4 ≤ x ≤ 3𝜋/4, cos(x) is positive and sin(x) is negative. Therefore, cos(x) - sin(x) > 0 in this interval. In the interval 3𝜋/4 ≤ x ≤ 5𝜋/4, both cos(x) and sin(x) are negative. Therefore, cos(x) - sin(x) < 0 in this interval. In the interval 5𝜋/4 ≤ x ≤ 7𝜋/4, cos(x) is negative and sin(x) is positive. Therefore, cos(x) - sin(x) < 0 in this interval. In the interval 7𝜋/4 ≤ x ≤ 2𝜋, both cos(x) and sin(x) are positive. Therefore, cos(x) - sin(x) > 0 in this interval. Combining all the intervals where cos(x) - sin(x) < 0, we have: 𝜋/4 ≤ x ≤ 3𝜋/4 ∪ 5𝜋/4 ≤ x ≤ 7𝜋/4 So, the interval(s) on which f is decreasing is [𝜋/4, 3𝜋/4] ∪ [5𝜋/4, 7𝜋/4]. (c) To find the local minimum and maximum values of f, we need to find the critical points of f(x) and evaluate the function at those points. The critical points occur when the derivative f'(x) is equal to zero or undefined. Let's find the critical points by setting f'(x) = 0: 9 cos(x) - 9 sin(x) = 0 Dividing both sides by 9, we get: cos(x) - sin(x) = 0 Simplifying further, we have: cos(x) = sin(x) This equation is satisfied when x = 𝜋/4 or x = 5𝜋/4. Now, let's evaluate f(x) at these critical points and the endpoints of the interval [0, 2𝜋]: f(0) = 9 sin(0) + 9 cos(0) = 9(0) + 9(1) = 9 f(𝜋/4) = 9 sin(𝜋/4) + 9 cos(𝜋/4) = 9(1/√2) + 9(1/√2) = 9√2 f(3𝜋/4) = 9 sin(3𝜋/4) + 9 cos(3𝜋/4) = 9(1/√2) - 9(1/√2) = 0 f(2𝜋) = 9 sin(2𝜋) + 9 cos(2𝜋) = 9(0) + 9(1) = 9 From these evaluations, we can see that the local minimum value of f is 0 and the local maximum value of f is 9√2.

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(a) To find the interval(s) on which f is increasing, we need to determine where the derivative of f is positive. Let's find the derivative of f(x) first. f'(x) = 9 cos(x) - 9 sin(x) To find where f'(x) > 0, we need to solve the inequality: 9 cos(x) - 9 sin(x) > 0 Dividing both sides by 9, we get: cos(x) - sin(x) > 0 Now, we can analyze the sign of cos(x) - sin(x) in different intervals. For 0 ≤ x ≤ 2𝜋, we can divide this interval into smaller intervals based on the behavior of cos(x) and sin(x). In the interval 0 ≤ x ≤ 𝜋/4, both cos(x) and sin(x) are positive. Therefore, cos(x) - sin(x) > 0 in this interval. In the interval 𝜋/4 ≤ x ≤ 3𝜋/4, cos(x) is positive and sin(x) is negative. Therefore, cos(x) - sin(x) > 0 in this interval. In the interval 3𝜋/4 ≤ x ≤ 5𝜋/4, both cos(x) and sin(x) are negative. Therefore, cos(x) - sin(x) < 0 in this interval. In the interval 5𝜋/4 ≤ x ≤ 7𝜋/4, cos(x) is negative and sin(x) is positive. Therefore, cos(x) - sin(x) < 0 in this interval. In the interval 7𝜋/4 ≤ x ≤ 2𝜋, both cos(x) and sin(x) are positive. Therefore, cos(x) - sin(x) > 0 in this interval. Combining all the intervals where cos(x) - sin(x) > 0, we have: 0 ≤ x ≤ 𝜋/4 ∪ 𝜋/4 ≤ x ≤ 3𝜋/4 ∪ 7𝜋/4 ≤ x ≤ 2𝜋 So, the interval(s) on which f is increasing is [0, 𝜋/4] ∪ [𝜋/4, 3𝜋/4] ∪ [7𝜋/4, 2𝜋]. (b) To find the interval(s) on which f is decreasing, we need to determine where the derivative of f is negative. Let's find the derivative of f(x) again. f'(x) = 9 cos(x) - 9 sin(x) To find where f'(x) < 0, we need to solve the inequality: 9 cos(x) - 9 sin(x) < 0 Dividing both sides by 9, we get: cos(x) - sin(x) < 0 Now, we can analyze the sign of cos(x) - sin(x) in different intervals. For 0 ≤ x ≤ 2𝜋, we can divide this interval into smaller intervals based on the behavior of cos(x) and sin(x). In the interval 0 ≤ x ≤ 𝜋/4, both cos(x) and sin(x) are positive. Therefore, cos(x) - sin(x) > 0 in this interval. In the interval 𝜋/4 ≤ x ≤ 3𝜋/4, cos(x) is positive and sin(x) is negative. Therefore, cos(x) - sin(x) > 0 in this interval. In the interval 3𝜋/4 ≤ x ≤ 5𝜋/4, both cos(x) and sin(x) are negative. Therefore, cos(x) - sin(x) < 0 in this interval. In the interval 5𝜋/4 ≤ x ≤ 7𝜋/4, cos(x) is negative and sin(x) is positive. Therefore, cos(x) - sin(x) < 0 in this interval. In the interval 7𝜋/4 ≤ x ≤ 2𝜋, both cos(x) and sin(x) are positive. Therefore, cos(x) - sin(x) > 0 in this interval. Combining all the intervals where cos(x) - sin(x) < 0, we have: 𝜋/4 ≤ x ≤ 3𝜋/4 ∪ 5𝜋/4 ≤ x ≤ 7𝜋/4 So, the interval(s) on which f is decreasing is [𝜋/4, 3𝜋/4] ∪ [5𝜋/4, 7𝜋/4]. (c) To find the local minimum and maximum values of f, we need to find the critical points of f(x) and evaluate the function at those points. The critical points occur when the derivative f'(x) is equal to zero or undefined. Let's find the critical points by setting f'(x) = 0: 9 cos(x) - 9 sin(x) = 0 Dividing both sides by 9, we get: cos(x) - sin(x) = 0 Simplifying further, we have: cos(x) = sin(x) This equation is satisfied when x = 𝜋/4 or x = 5𝜋/4. Now, let's evaluate f(x) at these critical points and the endpoints of the interval [0, 2𝜋]: f(0) = 9 sin(0) + 9 cos(0) = 9(0) + 9(1) = 9 f(𝜋/4) = 9 sin(𝜋/4) + 9 cos(𝜋/4) = 9(1/√2) + 9(1/√2) = 9√2 f(3𝜋/4) = 9 sin(3𝜋/4) + 9 cos(3𝜋/4) = 9(1/√2) - 9(1/√2) = 0 f(2𝜋) = 9 sin(2𝜋) + 9 cos(2𝜋) = 9(0) + 9(1) = 9 From these evaluations, we can see that the local minimum value of f is 0 and the local maximum value of f is 9√2.

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