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burhan101
Determine and classify all critical points of each function on the interval -4<x<4 f(x) = x⁴-4x³
Critical points are found via taking the first derivative of the function...and setting it = to 0 So what is \[\frac{ d }{ dx }x^4-4x^3\]
One step ahead of me...so you will have 2 critical points.. being...?
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
ok, between them, which one is bigger? the bigger is local max, the smaller is local min. right?
what do you mean which one :S
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?
@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 ?