anonymous
  • anonymous
A(t)=t^5-5t^4+20t^3-17 I am having trouble setting the derivative equal to zero. Please help!
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
What did you get for the derivative?
anonymous
  • anonymous
I've got \[A' = 5t^4-20t^3 + 60t^2 = 5t^2(t^2-4t + 12)\]
anonymous
  • anonymous
THat's exactly what I've got

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myininaya
  • myininaya
a*b=0 implies either a=0 or b=0 or both=0
anonymous
  • anonymous
Well I know that 5t^2=0 but I when I solve the parentheses for 0 (using quadratic formula) it comes out negative. I can't fit that into the rest of my problem!
anonymous
  • anonymous
(I am solving for the intervals of increasing and decreasing in the function)
myininaya
  • myininaya
you mean the number under the square root is negative?
anonymous
  • anonymous
Yes
myininaya
  • myininaya
so it is imaginary that means your only critical number is 0
anonymous
  • anonymous
That just means that that part of the equation doesn't have any real roots.
anonymous
  • anonymous
Okay. So if I am solving for the intervals, do I just divide my number in half?
myininaya
  • myininaya
I like using test points before and after each critical number
anonymous
  • anonymous
You're looking for intervals for what? Where it's increasing and decreasing?
anonymous
  • anonymous
Yes, @polpak
myininaya
  • myininaya
the derivative tells us the slope therefore tells us if is increasing or decreasing so choose a number before and after 0 to see what if it is increasing or decreasing
anonymous
  • anonymous
So it's increasing from negative infinity to infinity, correct?
anonymous
  • anonymous
The derivative is always positive except at 0. So it is increasing for \(x \in (-\infty,0) \bigcup (0,\infty)\)
anonymous
  • anonymous
It is not increasing at 0.
myininaya
  • myininaya
it has a horizontal tangent at x=0 thats what we found when we set A'=0
myininaya
  • myininaya
A'=0 means we are trying to find all x that has slope 0
anonymous
  • anonymous
Okay, thanks guys!

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