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Determine and classify all critical points of each function on the interval -4

Mathematics
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Critical points are found via taking the first derivative of the function...and setting it = to 0 So what is \[\frac{ d }{ dx }x^4-4x^3\]
0=4x²(x-3)
One step ahead of me...so you will have 2 critical points.. being...?

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Other answers:

0 and 3 ?
That would be correct...Now do you have to classify if they are a max or a min or an inflection point?
How do i find that out?
plug them into the original function and calculate its value. f(0) =? and f(3)=? compare them to give out the answer
(3, -27)
ok, between them, which one is bigger? the bigger is local max, the smaller is local min. right?
yes
what do you mean which one :S
like between (x,y)
for your problem, so far you have 4 critical points, -4, 0, 3, 4, just plug them into the original function and make conclusion. done.
where did -4 and 4 come from ?
$$f(x) = x^4-4x^3\\f'(x)=4x^3-12x^2=4x^2(x-3)\\4x^2(x-3)=0\implies4x^2=0,x-3=0$$... so we conclude \(x=0\) and \(x=3\) are our critical points. Both lie in our interval so classify them.
@burhan101 they're our endpoints, which you'd want to test when finding *absolute* extrema. @Loser66 had a minor misunderstanding
@oldrin.bataku what do i do with these points now?
plug them into f(x)?
@burhan101 no you want to check the second derivative:$$f''(x)=12x^2-24x=12x(x-2)\\f''(0)=0\\f''(3)=12(3)=36>0$$... so our derivative is increasing near \(x=3\) meaning \(x=3\) is a relative minimum; since our derivative is neither increasing nor decreasing near \(x=0\) we find it's neither! (hence the even multiplicity)
why are we plugging in 3 ?

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