## anonymous 3 years ago Determine and classify all critical points of each function on the interval -4<x<4 f(x) = x⁴-4x³

1. johnweldon1993

Critical points are found via taking the first derivative of the function...and setting it = to 0 So what is $\frac{ d }{ dx }x^4-4x^3$

2. anonymous

0=4x²(x-3)

3. johnweldon1993

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

4. anonymous

0 and 3 ?

5. johnweldon1993

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

6. anonymous

How do i find that out?

7. Loser66

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

8. anonymous

(3, -27)

9. Loser66

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

10. anonymous

yes

11. anonymous

what do you mean which one :S

12. anonymous

like between (x,y)

13. Loser66

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.

14. anonymous

where did -4 and 4 come from ?

15. anonymous

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

16. anonymous

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

17. anonymous

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

18. anonymous

plug them into f(x)?

19. anonymous

@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)

20. anonymous

why are we plugging in 3 ?