$\frac{(x+11)(x-13)}{x+9} \leq 0$
Answer (2)
Find the critical points of the inequality: x = − 11 , x = − 9 , and x = 13 .
Create intervals based on the critical points: ( − ∞ , − 11 ] , [ − 11 , − 9 ) , ( − 9 , 13 ] , and [ 13 , ∞ ) .
Test each interval to determine where the inequality holds: ( − ∞ , − 11 ] and ( − 9 , 13 ] .
Combine the intervals to find the solution set: ( − ∞ , − 11 ] ∪ ( − 9 , 13 ] . The final answer is ( − ∞ , − 11 ] ∪ ( − 9 , 13 ] .
Explanation
Understanding the Problem We are given the inequality x + 9 ( x + 11 ) ( x − 13 ) ≤ 0 . Our goal is to find all values of x that satisfy this inequality. To do this, we will first identify the critical points, which are the values of x that make the numerator or the denominator equal to zero. These points divide the number line into intervals, and we will test each interval to see if the inequality holds.
Finding Critical Points The critical points are found by setting the numerator and denominator equal to zero:
Numerator: ( x + 11 ) ( x − 13 ) = 0 , which gives x = − 11 and x = 13 .
Denominator: x + 9 = 0 , which gives x = − 9 .
So, the critical points are x = − 11 , x = − 9 , and x = 13 . Note that x = − 9 makes the denominator zero, so the function is undefined at this point.
Creating Intervals Now, we create a number line and mark the critical points: − 11 , − 9 , and 13 . These points divide the number line into four intervals: ( − ∞ , − 11 ] , [ − 11 , − 9 ) , ( − 9 , 13 ] , and [ 13 , ∞ ) . We will test a value from each interval to see if the inequality x + 9 ( x + 11 ) ( x − 13 ) ≤ 0 holds.
Testing Intervals
Interval ( − ∞ , − 11 ] : Choose a test point, say x = − 12 . Then − 12 + 9 ( − 12 + 11 ) ( − 12 − 13 ) = − 3 ( − 1 ) ( − 25 ) = − 3 25 ≤ 0. So, this interval is part of the solution.
Interval [ − 11 , − 9 ) : Choose a test point, say x = − 10 . Then 0."> − 10 + 9 ( − 10 + 11 ) ( − 10 − 13 ) = − 1 ( 1 ) ( − 23 ) = 23 > 0. So, this interval is not part of the solution.
Interval ( − 9 , 13 ] : Choose a test point, say x = 0 . Then 0 + 9 ( 0 + 11 ) ( 0 − 13 ) = 9 ( 11 ) ( − 13 ) = − 9 143 ≤ 0. So, this interval is part of the solution.
Interval [ 13 , ∞ ) : Choose a test point, say x = 14 . Then 0."> 14 + 9 ( 14 + 11 ) ( 14 − 13 ) = 23 ( 25 ) ( 1 ) = 23 25 > 0. So, this interval is not part of the solution.
Final Solution Combining the intervals where the inequality holds, we have ( − ∞ , − 11 ] and ( − 9 , 13 ] . Since x = − 9 , the solution is ( − ∞ , − 11 ] ∪ ( − 9 , 13 ] .
Examples
Understanding inequalities like this is crucial in many real-world scenarios. For instance, in economics, you might use inequalities to determine the range of prices for a product that ensures profitability. If you know your cost function and revenue function, setting up an inequality can help you find the price points that keep your business in the black. Similarly, in engineering, inequalities are used to set tolerance levels for designs, ensuring that components function correctly within specified ranges.
To solve the inequality x + 9 ( x + 11 ) ( x − 13 ) ≤ 0 , we find critical points at x = − 11 , x = − 9 , and x = 13 . Testing intervals based on these points reveals the solution is ( − ∞ , − 11 ] ∪ ( − 9 , 13 ] .
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