Suppose f(t) = (t-4)(t+9)
A.) for which values of t is the function f(t) positive? Enter using inequalities.
B.) for which values of t is the function f(x) negative? Enter your answer using inequalities.
+773 Nota bene: positive means greater than 0, and negative means less than 0. Positive and negative real numbers do not include 0. If you wanted non-positive and non-negative, you need to restate your question. Otherwise, here is the solution.
A) To have f(t) be positive, either \((t-4)\) and \((t+9)\) must both be positive or \((t-4)\) and \((t+9)\) must both be negative. In order to have (t-4)(t+9) be positive, since (t-4) < (t+9), we need (t-4) to be greater than 0. We have the inequality \(t-4>0\), which simplifies into \(t > 4\). We cannot have t = 4 because (4-4) = 0, and anything multiplied to 0 is 0 (which is not positive). We can also plug in some values of \(t\) and try them out. If t = 5, we have \((5-4)\cdot(5+9)\), which simplifies into \((1)\cdot(14)\), which is greater than 0.
In order to have (t-4) and (t+9) to both be negative, since (t+9) > (t-4), we need (t+9) to be less than 0. We have the inequality \(t+9<0\), which simplifies into \(t<-9\). We cannot have t = -9 because (-9+9) = 0, and anything multiplied by 0 is 0. If we try t = -10, we have \((-10-4)\cdot(-10+9)\), which simplifies into \((-14)\cdot(-1)\). Since the minus signs cancel out each other, we are left with 14, which is greater than 0.
Combining both inequalities \(t>4\) and \(t<-9\), we have the answer: \(\boxed{t<-9}\) or \(\boxed{t>4}\) (or \(\boxed{t \in (-\infty, -9)\cup(4, \infty)}\)in intervals).
B) To have f(t) be negative, either \((t-4)\) is positive and \((t+9)\) is negative or \((t-4)\) is negative and \((t+9)\) is positive. If (t-4) is positive, then (t+9) must also be positive because (t+9) > (t-4), so we can eliminate the first scenario. Since (t+9) > (t-4), to have (t+9) be positive, \(t\) must be more than -9. We cannot have t = -9 because (-9+9) = 0, and anything multiplied to 0 is 0. \(t\) can also not be more than 4 because (t-4) would then equal 0 or be positive, which is not what we want. Combining the inequalities \(t>-9\) and \(t<4\), we have the answer: \(\boxed{-9 .
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- PartialMathematician
+773 Nota bene: positive means greater than 0, and negative means less than 0. Positive and negative real numbers do not include 0. If you wanted non-positive and non-negative, you need to restate your question. Otherwise, here is the solution.
A) To have f(t) be positive, either \((t-4)\) and \((t+9)\) must both be positive or \((t-4)\) and \((t+9)\) must both be negative. In order to have (t-4)(t+9) be positive, since (t-4) < (t+9), we need (t-4) to be greater than 0. We have the inequality \(t-4>0\), which simplifies into \(t > 4\). We cannot have t = 4 because (4-4) = 0, and anything multiplied to 0 is 0 (which is not positive). We can also plug in some values of \(t\) and try them out. If t = 5, we have \((5-4)\cdot(5+9)\), which simplifies into \((1)\cdot(14)\), which is greater than 0.
In order to have (t-4) and (t+9) to both be negative, since (t+9) > (t-4), we need (t+9) to be less than 0. We have the inequality \(t+9<0\), which simplifies into \(t<-9\). We cannot have t = -9 because (-9+9) = 0, and anything multiplied by 0 is 0. If we try t = -10, we have \((-10-4)\cdot(-10+9)\), which simplifies into \((-14)\cdot(-1)\). Since the minus signs cancel out each other, we are left with 14, which is greater than 0.
Combining both inequalities \(t>4\) and \(t<-9\), we have the answer: \(\boxed{t<-9}\) or \(\boxed{t>4}\) (or \(\boxed{t \in (-\infty, -9)\cup(4, \infty)}\)in intervals).
B) To have f(t) be negative, either \((t-4)\) is positive and \((t+9)\) is negative or \((t-4)\) is negative and \((t+9)\) is positive. If (t-4) is positive, then (t+9) must also be positive because (t+9) > (t-4), so we can eliminate the first scenario. Since (t+9) > (t-4), to have (t+9) be positive, \(t\) must be more than -9. We cannot have t = -9 because (-9+9) = 0, and anything multiplied to 0 is 0. \(t\) can also not be more than 4 because (t-4) would then equal 0 or be positive, which is not what we want. Combining the inequalities \(t>-9\) and \(t<4\), we have the answer: \(\boxed{-9 .
Hope this helps, Guest! (If you sign in or register, your questions will be answered faster and will have a higher priority than guests)
- PartialMathematician
+773 How long are these answers supposed to be, CPhill and Melody? I spent over 20 minutes writing this much. I am not sure what the average time spent answering a difficult question is, so I am going to assume 20 minutes is not too long. Thank you for clarifying. 
- PartialMathematician
+1253 For me, if it's an easier problem I'll take less time and harder problems I'll take more. Usually, 5 minutes for the easy problems (e.g. Solve 2(x-3)+6=5) and 10-15 minutes for the harder problems (e.g. some type of geometry question). The most I've taken in a problem is, say, 20-25 minutes.
+128475 Thanks for that answer, PM....
The answers don't have any "space limits" ..........as long as you're giving good answers [ or.....even attempting a difficult problem..... ]
However......don't write a book.....the person asking the question will probably not appreciate extremely long answers
