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What did you get for the derivative?
I've got \[A' = 5t^4-20t^3 + 60t^2 = 5t^2(t^2-4t + 12)\]
THat's exactly what I've got
a*b=0 implies either a=0 or b=0 or both=0
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!
(I am solving for the intervals of increasing and decreasing in the function)
you mean the number under the square root is negative?
so it is imaginary that means your only critical number is 0
That just means that that part of the equation doesn't have any real roots.
Okay. So if I am solving for the intervals, do I just divide my number in half?
I like using test points before and after each critical number
You're looking for intervals for what? Where it's increasing and decreasing?
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
So it's increasing from negative infinity to infinity, correct?
The derivative is always positive except at 0. So it is increasing for \(x \in (-\infty,0) \bigcup (0,\infty)\)
It is not increasing at 0.
it has a horizontal tangent at x=0 thats what we found when we set A'=0
A'=0 means we are trying to find all x that has slope 0
Okay, thanks guys!