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anonymous
 3 years ago
Precalc help please???:( Solving inequalities with one variable? Sign chart?
*In exercises 16, determine the x values that cause the polynomial function to be (a)zero (b)positive (c)negative.
anonymous
 3 years ago
Precalc help please???:( Solving inequalities with one variable? Sign chart? *In exercises 16, determine the x values that cause the polynomial function to be (a)zero (b)positive (c)negative.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[f()=(x+2)(x+1)(x5)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry it should be f(x)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0First you figure out when f(x) = 0, which occurs when (x+2) =0 or (x+1) =0 or (x5) =0. That is pretty easy, x=2,1,5. Your number line is (2)(1)(5) Pick a number left of (2) say (5) and input it in as f(5)= 120 therefore The part left of (2) is negative. The next number is between (2) and (1) and is therefore something like (1.5). f(1.5) = something positive because when you put it into (x+2) you get a positive number, and into the other two parts of the equation a negative number returns. (1)(1)(1) = 1. Therefore that part is positive. Doing so for the other spaces (1) to (5) and (5) to infinity results in a negative for (1) to (5) and a positive for (5) to infinity. Therefore: negative(2)positive(1)negative(5)positive

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay i understand that but im unsure of what to put for the answers, because in the back of the book (answers) their like this, (b) 2<x<1 or x>5 (c) x<2 or 1<x<5? but i have no clue how they got that

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That's what I get for not fully reading the question. They just want to know when that f(x) is positive, a zero, or negative. You use the sign chart to determine when it is negative and positive. So for (a) you wish to find the zeros which are at (2), (1), and (5). (b) however asks for when the equation is positive. It just wants the ranges. So the sign chart says that the equation is positive from (2) to (1) or x is greater than (2) but less than (1). Therefore, 2<x<1. You don't use an equals inequality because that would imply that both (2) and (1) are positive and we know those to be zeros. The second part of the answer for (b) is the second part that is positive: 5<x and then onto infinity. (c) does the same thing but with negatives instead.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0omg thankyou so much i wouldve failed my test tomorrow lol my teacher was no help <333
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