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anonymous
 3 years ago
Determine and classify all critical points of each function on the interval 4<x<4
f(x) = x⁴4x³
anonymous
 3 years ago
Determine and classify all critical points of each function on the interval 4<x<4 f(x) = x⁴4x³

This Question is Closed

johnweldon1993
 3 years ago
Best ResponseYou've already chosen the best response.0Critical points are found via taking the first derivative of the function...and setting it = to 0 So what is \[\frac{ d }{ dx }x^44x^3\]

johnweldon1993
 3 years ago
Best ResponseYou've already chosen the best response.0One step ahead of me...so you will have 2 critical points.. being...?

johnweldon1993
 3 years ago
Best ResponseYou've already chosen the best response.0That would be correct...Now do you have to classify if they are a max or a min or an inflection point?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0How do i find that out?

Loser66
 3 years ago
Best ResponseYou've already chosen the best response.0plug them into the original function and calculate its value. f(0) =? and f(3)=? compare them to give out the answer

Loser66
 3 years ago
Best ResponseYou've already chosen the best response.0ok, between them, which one is bigger? the bigger is local max, the smaller is local min. right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what do you mean which one :S

Loser66
 3 years ago
Best ResponseYou've already chosen the best response.0for your problem, so far you have 4 critical points, 4, 0, 3, 4, just plug them into the original function and make conclusion. done.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0where did 4 and 4 come from ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0$$f(x) = x^44x^3\\f'(x)=4x^312x^2=4x^2(x3)\\4x^2(x3)=0\implies4x^2=0,x3=0$$... so we conclude \(x=0\) and \(x=3\) are our critical points. Both lie in our interval so classify them.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@burhan101 they're our endpoints, which you'd want to test when finding *absolute* extrema. @Loser66 had a minor misunderstanding

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@oldrin.bataku what do i do with these points now?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@burhan101 no you want to check the second derivative:$$f''(x)=12x^224x=12x(x2)\\f''(0)=0\\f''(3)=12(3)=36>0$$... 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)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0why are we plugging in 3 ?
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