If $f(x)=3x^y, y∈N$ what should the value be for $y$ so that their is a maxima or minima at $x=0$?

Question

If $f(x)=3x^y, y∈N$ what should the value be for $y$ so that their is a maxima or minima at $x=0$? So far I've done trial and error, with $y=3, 4,5$ but for every $n$ value, you'd have to find the derivative, so for $y=3$ $f'(x)=6x^2$. When you make $f'(x)=0$ you get $x=0$ and then I found that for all $y$ value you have $f''(0)=0$ and not $f''(0)>0$ or $f''(0)<0$. At this point I'm not sure if there even is a maxima or minima at $x=0$

Answer 1

Note that for $y=1\implies f(x)=3x$.

For y>1

• if $y$ odd

$$f'(x)=3yx^{y-1}\geq0 \quad f'(x)=0\iff x=0$$

thus $f(x)$ is strictly increasing and $x=0$ is an inflection point (EG $f(x)=3x^3$).

• if $y$ even

$$f'(x)=3yx^{y-1}>0 \quad x>0$$

$$f'(x)=3yx^{y-1}<0 \quad x<0$$

$$f'(x)=0\iff x=0$$

thus $f(x)$ is a minimum point (EG $f(x)=3x^4$).

Answer 2

I would suggest a "much more basic" approach to this question: Let $y \in \mathbb{N}$.

Consider as a first case: $y$ odd: Then $f(x) = 3x^{2k+1}$ with $k \in \mathbb{N}$. With this $f(x) = 3x \underbrace{x^{2k}}_{\geq 0}$ and therefore $0$ cannot be a minimum or a maximum, since $f$ has a change of sign in $0$.

Secondly for $y$ even, follow the same argumentation and observe $f(0)=0$.