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What'd you come up with? It's a pretty straight forward derivative to take.
what i am wondering is if i should take an x out first. x(3x^3 - 2x^2 - 12x +18)
Oh, no. Take the derivative term by term. It'll be much simplier.
In general I find it's easier to multiply out the products and take the derivative term by term unless it's a particularly hairy product with a lot of exponents, etc.
ok, the whole problem is: Find all critical point of: (the orginal equation i entered) and use the second derivative test to classify each as a relative man and rel min or neither. it seems to be more difficult to get my critical points.
\[f' = 12x^3 -6x^2 - 24x + 18\] \[f' = 0 \implies 2x^3 - x^2 - 4x + 3 = 0\]
when you say taking the derivative, I am assuming you mean the derivative with respect to x? Therefore, the derivative would symply be: f'(x)=12x^3-6x^2-24x+18
should have been relative max and rel min
with having so many x's i am not sure how to find the critical points
Yeah it's a tough one to factor.
should i pull an x out before i get the derivative
We can try..
I'm not sure it'll make it better.
i had done that already and it still has alot of x's
It pretty much gives the same result. I was hoping it would be of a form that would better suggest how to factor.
\(2x^3−x^2−4x+3=0\) \(x^2(2x-1) -(2x - 1) - (2x -1) -1\)
Bleh. I need to practice dividing polynomials more apparently ;p
WA claims it's \((x-1)^2(2x+3)\)
thanks for your help. i will work on this again tomorrow
one more quick question. should i go ahead and get the second derivative before i can get the critical points.
i really dont think i will. i think i get the critical points from the first derivative and then i get the second derivative to test.