Question

why f(x) = x^3 doesn't have relative extrema at 0

Answer 1

f'(x)=3x^2
first derivative test...now let's test some values around the critical point f'(x)=0
3x^2=0
x=0
\[(-\infty, 0), (0, \infty)\]
f'(1)=3
f'(-1)=3
No sign change here, no relative extrema

Answer 2

whereas f(x) = x^2 have rel extrema at 0

Answer 3

got it thanxxxxxxxx

Answer 4

f(x)=x^2
f'(x)=2x
2x=0
critical point x=0
\[(-\infty, 0), (0, \infty)\]
f'(-1)=-2
f'(1)=2
negative to positive
==>> local min

Answer 5

Increasing to Decreasing menas we have a local max
Decreasing to Increasing measn we have a local min
In other words you have to have a switch!
f=x^3 is increasing from (-inf,inf) there is no local extrema(or abs extrema)
f=x^2 has local min since we have the switch is from decreasing to increasing

Answer 6

on (-inf,inf)*