The derivative of the function \( f \) is defined by \( f^{\prime}(x)=x^{2}+4 \cos (2 x+2) \) for \( -2
Ask by Lewis Hardy. in the United States
Jan 27,2025
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To determine where the function \( f \) is increasing, we need to find the intervals where its derivative \( f^{\prime}(x) = x^{2} + 4 \cos(2x + 2) \) is greater than zero. 1. First, we analyze the behavior of the derivative. The term \( x^{2} \) is always non-negative and increases as \( x \) increases. The cosine term oscillates between -1 and 1, which means \( 4 \cos(2x + 2) \) oscillates between -4 and 4. Thus, the minimum value of the derivative occurs when \( \cos(2x + 2) = -1 \), giving us \( f^{\prime}(x) = x^{2} - 4 \). 2. To find where \( f^{\prime}(x) > 0 \): \[ x^{2} - 4 > 0 \implies x^{2} > 4 \implies x < -2 \text{ or } x > 2 \] However, we are limited to the domain \( -2 < x < 4 \), so only \( x > 2 \) is relevant here. 3. Now, we need to further analyze the derivative in the interval \( -2 < x < 4 \), particularly near 2: - For \( -2 < x < 2 \): Check values (let's test \( x = 0 \)): \[ f^{\prime}(0) = 0^{2} + 4 \cos(2 \cdot 0 + 2) = 4 \cos(2) \approx 4 \cdot (-0.416) \approx -1.664 \quad (\text{negative}) \] - For \( 2 < x < 4 \): Check values (let's test \( x = 3 \)): \[ f^{\prime}(3) = 3^{2} + 4 \cos(2 \cdot 3 + 2) = 9 + 4 \cos(8) \approx 9 + 4 \cdot 0.990 \approx 9 + 3.960 = 12.960 \quad (\text{positive}) \] Thus, the function \( f \) is increasing on the interval \( (2, 4) \). In summary, the function \( f \) is increasing where \( f^{\prime}(x) > 0 \), specifically in the interval: **Answer: \( (2, 4) \)**