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Solve x(x- 5) 2(x+2)2>0 and write your solution using interval notation. Hint: x (x- 5)2(x+2) 2=0 when x=-2, 0, 5. So these are the only places that the function could change from positive to negative. Form the intervals (- 00,- 2) U (- 2,0) U (0,5) U (5, o0 ) and test a number from each interval in the inequality to determine if it is true or false. For example, you could use x=-3 for the first interval, x=-1 for the second interval, x=1 for the third interval, and x=6 for the fourth interval. You do not have to use these numbers but any number chosen in a given interval will produce the same result(either positive or negative). - 3(- 3-5)2(- 3+2)²>0? – 3(- 8) 2(– 1)2>0? If x=-3, - 3(64) (1) >0? - 192>0? Ac This is false, so the interval (-o0,-2) would not be true. If it were true, then (-00,-2) would be part of the solution.

Solve x(x- 5) 2(x+2)2>0 and write your solution using interval notation. Hint: x (x- 5)2(x+2) 2=0 when x=-2, 0, 5. So these are the only places that the function could change from positive to negative. Form the intervals (- 00,- 2) U (- 2,0) U (0,5) U (5, o0 ) and test a number from each interval in the inequality to determine if it is true or false. For example, you could use x=-3 for the first interval, x=-1 for the second interval, x=1 for the third interval, and x=6 for the fourth interval. You do not have to use these numbers but any number chosen in a given interval will produce the same result(either positive or negative). - 3(- 3-5)2(- 3+2)²>0? – 3(- 8) 2(– 1)2>0? If x=-3, - 3(64) (1) >0? - 192>0? Ac This is false, so the interval (-o0,-2) would not be true. If it were true, then (-00,-2) would be part of the solution.

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Solve x (x - 5)²(x+2)²>0 and write your solution using interval notation. Hint: x (x – 5)2(x+2)2=0 when x=-2, 0, 5. So these are the only places that the function could change from positive to negative. Form the intervals (- 0, - 2) U (- 2,0) U (0,5) U (5, 0 ) and test a number from each interval in the inequality to determine if it is true or false. For example, you could use x=-3 for the first interval, x=-1 for the second interval, x=1 for the third interval, and x=6 for the fourth interval. You do not have to use these numbers but any number chosen in a given interval will produce the same result(either positive or negative). -3(- 3-5)2(- 3+2)2>0? – 3(- 8)2(-1)²>0? If x=-3, - 3(64) (1) >0? - 192>0? Ac Go This is false, so the interval (-00,-2) would not be true. If it were true, then (-00,-2) would be part of the solution.
Transcribed Image Text:Solve x (x - 5)²(x+2)²>0 and write your solution using interval notation. Hint: x (x – 5)2(x+2)2=0 when x=-2, 0, 5. So these are the only places that the function could change from positive to negative. Form the intervals (- 0, - 2) U (- 2,0) U (0,5) U (5, 0 ) and test a number from each interval in the inequality to determine if it is true or false. For example, you could use x=-3 for the first interval, x=-1 for the second interval, x=1 for the third interval, and x=6 for the fourth interval. You do not have to use these numbers but any number chosen in a given interval will produce the same result(either positive or negative). -3(- 3-5)2(- 3+2)2>0? – 3(- 8)2(-1)²>0? If x=-3, - 3(64) (1) >0? - 192>0? Ac Go This is false, so the interval (-00,-2) would not be true. If it were true, then (-00,-2) would be part of the solution.
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