+0

# Suppose f(t) = (t-4)(t+9)

0
44
5

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.

Guest Nov 20, 2018

#1
+371
+2

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 PartialMathematician Nov 21, 2018 edited by PartialMathematician Nov 21, 2018 #1 +371 +2 Best Answer 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

PartialMathematician  Nov 21, 2018
edited by PartialMathematician  Nov 21, 2018
#2
+371
+2

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

PartialMathematician  Nov 21, 2018
#3
+357
+1

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.

CoolStuffYT  Nov 21, 2018
#4
+371
+2

PartialMathematician  Nov 21, 2018
#5
+92429
0

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

CPhill  Nov 21, 2018