Question

`f(x)=x^4` Have we critical point?

DOUBT about Increase, Decrease and critical point (Maximum and Minimum)
`f(x)=x^3` and `g(x)=x^4`
about the Analysis of Functions I: Increase, Decrease, Concavity, and point critic (Maximum and Minimum)

`f(x)=x^3`
1st derivative
`f'(x)=3x^2`
`->` `f'(x)=0`
`3x^2=0`
`x=0`
only increase

2nd derivative
`f''(x) = 6x`
`->` `f''(x)=0`
`6x = 0`
`x=0`
inflection point (0)

substitution root `x=0` from `f'(x)` in `f''(x)`
`f''(0) = 6.0 = 0`
about critical point we dont have

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The question is about `g(x)=x^4`

`f(x)=x^4`
1st derivative
`f'(x)=4x^3`
`->` `f'(x)=0`
root `x=0`
Increase: ]0, +∞[
Descrease: ]-∞,0[

2nd derivative
`f''(x) = 12x^2`
`->` `f''(x)=0`
`12x^2 = 0`
`x=0`
Only concave up

substitution `x=0` from `f'(x)` in `f''(x)`
`f''(0) = 12(0)^2 = 0`
THE RESULT IS ZERO TOO

Answer 1

Yes we have, see the proof below.

Explanation:

Let `f(x)=x^4`

Then

`f'(x)=4x^3`
`f''(x)=12x^2`
`f''(x)=24x`
`f^((iv))(x)=24`
so `f^((iv))(0)=24>0`
since the derivative of `f(x)` (at `x=0`) is of even order and positive we get
a minimum.
`P(0;0)`