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\( 6 x - 3 \leq 2 x - 11 \cos x \)

Ask by Garza Summers. in South Africa
Mar 13,2025

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The solution to the inequality \( 6x - 3 \leq 2x - 11 \cos x \) is all \( x \) in the interval \( (-2, -1) \) where the function \( f(x) = 4x + 11 \cos x - 3 \) is less than or equal to zero.

Solución

The inequality \( 6x - 3 \leq 2x - 11 \cos x \) can be rewritten as: \[ 6x - 2x + 11 \cos x - 3 \leq 0 \] This simplifies to: \[ 4x + 11 \cos x - 3 \leq 0 \] To solve this inequality, we need to find the values of \( x \) that satisfy it. However, since it involves both a linear term and a trigonometric term, we will analyze it graphically or numerically rather than algebraically. Let's define the function: \[ f(x) = 4x + 11 \cos x - 3 \] We will find the points where \( f(x) = 0 \) and then determine the intervals where \( f(x) \leq 0 \). Since I cannot directly solve this equation, I will evaluate \( f(x) \) at several points to find the roots and analyze the behavior of the function. Let's calculate \( f(x) \) at a few key points. Solve the equation by following steps: - step0: Solve for \(f\): \(f\times 0=4\times 0+11\cos\left(0\right)-3\) - step1: Multiply: \(f\times 0=0+11\cos\left(0\right)-3\) - step2: Multiply the numbers: \(0=8\) - step3: The statement is false: \(f \in \varnothing \) Solve the equation \( f(-1) = 4(-1) + 11 \cos(-1) - 3 \). Solve the equation by following steps: - step0: Solve for \(f\): \(f\left(-1\right)=4\left(-1\right)+11\cos\left(-1\right)-3\) - step1: Use the rules for multiplication and division: \(-f=-7+11\cos\left(1\right)\) - step2: Change the signs: \(f=7-11\cos\left(1\right)\) Solve the equation \( f(1) = 4(1) + 11 \cos(1) - 3 \). Solve the equation by following steps: - step0: Solve for \(f\): \(f\times 1=4\times 1+11\cos\left(1\right)-3\) - step1: Multiply: \(f=1+11\cos\left(1\right)\) It seems that I encountered issues while trying to solve the equation directly. However, I can still evaluate the function \( f(x) = 4x + 11 \cos x - 3 \) at specific points to understand its behavior. 1. **At \( x = 0 \)**: \[ f(0) = 4(0) + 11 \cos(0) - 3 = 11 - 3 = 8 \] 2. **At \( x = 1 \)**: \[ f(1) = 4(1) + 11 \cos(1) - 3 = 1 + 11 \cos(1) \] (We will calculate \( 11 \cos(1) \) numerically.) 3. **At \( x = -1 \)**: \[ f(-1) = 4(-1) + 11 \cos(-1) - 3 = -4 + 11 \cos(1) - 3 = 7 - 11 \cos(1) \] (We will also calculate \( 7 - 11 \cos(1) \) numerically.) Now, let's calculate the values of \( 11 \cos(1) \) and \( 7 - 11 \cos(1) \). Calculate the value by following steps: - step0: Calculate: \(11\cos\left(1\right)\) Calculate or simplify the expression \( 7 - 11*\cos(1) \). Calculate the value by following steps: - step0: Calculate: \(7-11\cos\left(1\right)\) - step1: Calculate the trigonometric value: \(1.056675\) Now we have the numerical evaluations: 1. **At \( x = 0 \)**: \[ f(0) = 8 \] 2. **At \( x = 1 \)**: \[ f(1) \approx 1 + 5.943325 \approx 6.943325 \] 3. **At \( x = -1 \)**: \[ f(-1) \approx 1.056675 \] ### Summary of Evaluations: - \( f(0) = 8 \) (positive) - \( f(1) \approx 6.943325 \) (positive) - \( f(-1) \approx 1.056675 \) (positive) Since \( f(x) \) is positive at these points, we need to check further values to find where \( f(x) \) might become zero or negative. Let's evaluate \( f(x) \) at \( x = -2 \) and \( x = 2 \) to see if we can find a sign change. Calculate the value by following steps: - step0: Calculate: \(4\times 2+11\cos\left(2\right)-3\) - step1: Multiply the numbers: \(8+11\cos\left(2\right)-3\) - step2: Subtract the numbers: \(5+11\cos\left(2\right)\) - step3: Calculate the trigonometric value: \(0.422385\) Calculate or simplify the expression \( 4*(-2) + 11*\cos(-2) - 3 \). Calculate the value by following steps: - step0: Calculate: \(4\left(-2\right)+11\cos\left(-2\right)-3\) - step1: Calculate: \(4\left(-2\right)+11\cos\left(2\right)-3\) - step2: Multiply the numbers: \(-8+11\cos\left(2\right)-3\) - step3: Subtract the numbers: \(-11+11\cos\left(2\right)\) - step4: Calculate the trigonometric value: \(-15.577615\) Now we have evaluated the function at two additional points: 1. **At \( x = -2 \)**: \[ f(-2) \approx -15.577615 \quad (\text{negative}) \] 2. **At \( x = 2 \)**: \[ f(2) \approx 0.422385 \quad (\text{positive}) \] ### Summary of Evaluations: - \( f(0) = 8 \) (positive) - \( f(1) \approx 6.943325 \) (positive) - \( f(-1) \approx 1.056675 \) (positive) - \( f(-2) \approx -15.577615 \) (negative) - \( f(2) \approx 0.422385 \) (positive) ### Conclusion: The function \( f(x) \) changes from positive to negative between \( x = -1 \) and \( x = -2 \), and it is positive again at \( x = 2 \). To find the exact intervals where \( f(x) \leq 0 \), we can conclude that there is at least one root between \( x = -2 \) and \( x = -1 \). To summarize the solution to the inequality \( 6x - 3 \leq 2x - 11 \cos x \): - The solution set includes values of \( x \) in the interval \( (-2, -1) \) where \( f(x) \leq 0 \). For a more precise solution, numerical methods or graphing would be required to find the exact roots.

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To tackle this inequality, let's start by rearranging it to one side for better clarity. You can rewrite it as: \[ 6x - 2x + 11 \cos x - 3 \leq 0 \] This simplifies to: \[ 4x + 11 \cos x - 3 \leq 0 \] Now, to approach solving the inequality, one might consider plotting the left side as a function of \( x \) to see where it is less than or equal to zero. The term \( 11 \cos x \) oscillates between -11 and 11, which introduces periodic behavior to the overall function. To find where \( 4x - 3 \leq -11 \) or \( 4x - 3 \geq 11 \), you can evaluate conditions on \( x \) and find the intersections with the oscillating cosine function. Finding critical points can lead to multiple solutions since cosine's periodic nature means it could very well cross our linear term in several locations! Always be sure to test your intervals to ensure you identify the complete range of solutions correctly.

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