anonymous
  • anonymous
F(t)=4t + 11 / 3t^2 + 5t +1 Will someone walk me through this or explain the process of finding the solution.
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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tkhunny
  • tkhunny
First, one might need to establish what you have written. I see \(F(t) = 4t + \dfrac{11}{3t^{2}} + 5t + 1\). If this is not as you intended to write, perhaps you could use parentheses to explain your intent?
anonymous
  • anonymous
agree with tkhunny cause I was lookin at F(t)=(4t + 11) / (3t^2 + 5t +1) and it was making me unhappy since the denominator won't factor
anonymous
  • anonymous
@kantalope That is how the problem is set up, I'm sorry for the misinterpretation.

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anonymous
  • anonymous
so did tkhunny write it correct or is it how you wrote it correct?
anonymous
  • anonymous
F(t)=(4t + 11) / (3t^2 + 5t +1)
tkhunny
  • tkhunny
Excellent use of parentheses. The minimum value of 3t^2 + 5t + 1 is what?
anonymous
  • anonymous
0?
tkhunny
  • tkhunny
?? Why do you thing that?
anonymous
  • anonymous
Well one of the way we were thought to find t was to equal t to 0 and work backwards but that was confusing then he showed us an other way which was even more confusing.
anonymous
  • anonymous
taught
tkhunny
  • tkhunny
This is a little different. As a rational function, we are really only concered where the Numerator is zero UNLESS (a really big unless) the denominator also happens to be zero. Then it's a different story. Thus my question. If you could have told be that the denominator NEVER is zero, we can just ignore it. Something factor that never is zero has nothing to do with whether or not an entire expression is zero. Does that make any sense at all.
anonymous
  • anonymous
Yes that makes sense.
tkhunny
  • tkhunny
Then it is important to know exactly where the denominator is zero. It's just a quadratic. Let's see what you get.
tkhunny
  • tkhunny
Well, I need to go. When you find the two places where the denominator is zero, just write them down and remember them. After that, find where the numerator is zero. It's just linear, so this should take only a moment. Once you have t = -11/4 for the numerator, just look at what you have for the denominator. If neither of them is t = -11/4, you're done.
anonymous
  • anonymous
well following tkhunny's advice I plugged the denominator into the quadratic equation and got \[\frac{(-5\pm \sqrt{13}) }{6 }\]

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