Question

Determine and classify all critical points of each function on the interval -4 12 years ago

Answer 1

Critical points are found via taking the first derivative of the function...and setting it = to 0

So what is

\[\frac{ d }{ dx }x^4-4x^3\]

Answer 2

0=4x²(x-3)

Answer 3

One step ahead of me...so you will have 2 critical points..

being...?

Answer 4

0 and 3 ?

Answer 5

That would be correct...Now do you have to classify if they are a max or a min or an inflection point?

Answer 6

How do i find that out?

Answer 7

plug them into the original function and calculate its value. f(0) =? and f(3)=? compare them to give out the answer

Answer 8

(3, -27)

Answer 9

ok, between them, which one is bigger? the bigger is local max, the smaller is local min. right?

Answer 10

yes

Answer 11

what do you mean which one :S

Answer 12

like between (x,y)

Answer 13

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.

Answer 14

where did -4 and 4 come from ?

Answer 15

$$f(x) = x^4-4x^3\\f'(x)=4x^3-12x^2=4x^2(x-3)\\4x^2(x-3)=0\implies4x^2=0,x-3=0$$... so we conclude \(x=0\) and \(x=3\) are our critical points. Both lie in our interval so classify them.

Answer 16

@burhan101 they're our endpoints, which you'd want to test when finding *absolute* extrema. @Loser66 had a minor misunderstanding

Answer 17

@oldrin.bataku what do i do with these points now?

Answer 18

plug them into f(x)?

Answer 19

@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)

Answer 20

why are we plugging in 3 ?