Section 2.7 Question 44 (x + 12)/(x + 2) ≥ 3 Simplifying the rational inequality just a bit, I get (-2x + 6)/(x + 2) ≥ 0 Set denominator to zero and solve for x. I get x = -2. This means x cannot be -2 here. If x = -2, we get division by zero which is undefined. Now, solving -2x + 6)/(x + 2) ≥ 0 for x, I get the following: x ≤ 3. Plot x = -2 and x = 3 on the real number line and test each interval. <-------(-2)---------(3)---------> When x = -3, we get a false statement. As stated above, x cannot be -2. When x = 0, we get a true statement. When x = 3, we get a true statement. When x = 4, we get a false statement. The only solution interval I see is [0,3]. Here it is on the real number line: You say?
Can you break it down step by step? How did you end up with (-2, 3]? While you show me how this is done, I will also watch a few video lessons. I need to solve at least 3 more rational inequality problems from the current section. I cannot go beyond Section 2.7 without knowing this very important precalculus topic. I need to feel confident with rational inequality. Otherwise, I stop right here. P. S. Look for some Calculus 1 questions tonight.
(-2x + 6)/(x + 2) ≥ 0 if (-2x + 6) ≥ 0 => x<=3 if denominator (x + 2) ≥ 0 => x>=-2 ; since denominator cannot be = to 0, we use only x>-2 combine solutions: -2<x<=3
Thank you. I will do 46, 48, 50 and 52 for additional practice BUT only after watching a few video clips.
