Find the range of values:
(a) [tex]\frac{x-3}{x+3}\ \textgreater \ 0[/tex]
(c) [tex]\frac{2 x+1}{x-3}\ \textgreater \ 0[/tex]
Answer (2)
Find critical points for each inequality by setting the numerator and denominator to zero.
Test intervals on a number line to determine where the inequality holds.
Express the solution as a union of intervals.
The solution to (a) is x < − 3 or 3"> x > 3 , and the solution to (c) is x < − 2 1 or 3"> x > 3 .
3 \text{ for (a), and } x < -\frac{1}{2} \text{ or } x > 3 \text{ for (c)}}"> x < − 3 or x > 3 for (a), and x < − 2 1 or x > 3 for (c)
Explanation
Problem Analysis We are given two inequalities to solve:
(a) 0"> x + 3 x − 3 > 0
(c) 0"> x − 3 2 x + 1 > 0
We need to find the range of values of x for which each inequality holds.
Solving Inequality (a) For inequality (a), 0"> x + 3 x − 3 > 0 , we find the critical points by setting the numerator and denominator equal to zero:
x − 3 = 0 ⇒ x = 3
x + 3 = 0 ⇒ x = − 3
These critical points divide the number line into three intervals: ( − ∞ , − 3 ) , ( − 3 , 3 ) , and ( 3 , ∞ ) . We test a value from each interval to determine where the inequality holds.
Interval ( − ∞ , − 3 ) : Let x = − 4 . Then 0"> − 4 + 3 − 4 − 3 = − 1 − 7 = 7 > 0 . So the inequality holds in this interval.
Interval ( − 3 , 3 ) : Let x = 0 . Then 0 + 3 0 − 3 = 3 − 3 = − 1 < 0 . So the inequality does not hold in this interval.
Interval ( 3 , ∞ ) : Let x = 4 . Then 0"> 4 + 3 4 − 3 = 7 1 > 0 . So the inequality holds in this interval.
Thus, the solution to inequality (a) is x < − 3 or 3"> x > 3 .
Solving Inequality (c) For inequality (c), 0"> x − 3 2 x + 1 > 0 , we find the critical points by setting the numerator and denominator equal to zero:
2 x + 1 = 0 ⇒ x = − 2 1
x − 3 = 0 ⇒ x = 3
These critical points divide the number line into three intervals: ( − ∞ , − 2 1 ) , ( − 2 1 , 3 ) , and ( 3 , ∞ ) . We test a value from each interval to determine where the inequality holds.
Interval ( − ∞ , − 2 1 ) : Let x = − 1 . Then 0"> − 1 − 3 2 ( − 1 ) + 1 = − 4 − 1 = 4 1 > 0 . So the inequality holds in this interval.
Interval ( − 2 1 , 3 ) : Let x = 0 . Then 0 − 3 2 ( 0 ) + 1 = − 3 1 = − 3 1 < 0 . So the inequality does not hold in this interval.
Interval ( 3 , ∞ ) : Let x = 4 . Then 0"> 4 − 3 2 ( 4 ) + 1 = 1 9 = 9 > 0 . So the inequality holds in this interval.
Thus, the solution to inequality (c) is x < − 2 1 or 3"> x > 3 .
Final Answer The solution to inequality (a) is x < − 3 or 3"> x > 3 , which can be written as ( − ∞ , − 3 ) ∪ ( 3 , ∞ ) .
The solution to inequality (c) is x < − 2 1 or 3"> x > 3 , which can be written as ( − ∞ , − 2 1 ) ∪ ( 3 , ∞ ) .
Examples
Understanding inequalities is crucial in various real-world scenarios. For instance, in economics, companies use inequalities to determine the price range that maximizes profit. Similarly, in engineering, inequalities are used to ensure that structures can withstand certain loads or stresses. In everyday life, we use inequalities when budgeting to ensure that our expenses do not exceed our income. These examples highlight the practical importance of mastering inequalities.
The solutions to the inequalities are: (a) x < − 3 or 3"> x > 3 , and (c) x < − 2 1 or 3"> x > 3 .
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