Find all local maximum points, local minimum points, and points of inflection on the curve y=-3x^{4}-8x^{3}+18x^{2}
What I did: First derivative is y'=-12x^{3}-24x^{2}+36x
Then I set it equal to 0
Then I factored it out: -12x(x^{2}+2x-3) = -12x(x-1)(x-2), so the critical numbers are x=0, x=1, and x=2
After getting the critical numbers, I did the second derivative
y''=36x^{2}-58x+36
I used the second derivative test, and they end up all positive, so it's all min,
So I'm a bit confused.
Isn't this the way to find absolute minimum, instead of local minimum o.o
And how do I go on from there
Oh oops, I factored it wrong
It's actually 12x(x+3)(x-1)
So x=0, -3, and 1
So the 2nd derative would be:
-36x^{2}-58x+36
Plugging in 0 would make it 36, (So Minimum)
Plugging in 1 would make it Negative (So Maximum)
Plugging in -3 would make it Positive (So Minimum)
Though, I'm a bit confused in the multiple choice answers, because x=-3, 1 is supposed to match up into a max or min, while x=0 is alone either a max or min.
If I use the factored one: -2(18x^{2}+24x-18)
(Or maybe the multiple choice answers could be wrong as well)
Points of inflection would be where 2nd derivative is equal to 0 right?
So I did the quadratic formula within the paranthesis, and I got something different: (-29+ or - sqrt2137)/36
While in the multiple choice answers, it was like either (-2 + or - sqrt13)/3 or (-2 + or - sqrt19)/3