Find ỉnteger values of $m$ such that the function: $f(x)=x^8 + (m - 2)x^5 - (m^2 - 4)x^4 + 1$ reaches local minimum at $x=0$
This is what I did:
We have: $f'(x)=8x^7+5(m-2)x^4+4(4-m^2)x^3=x^3\underbrace{[8x^4+(5m-10)x-4m^2+16]}_{g(x)}$
Let $\displaystyle{f'(x)=0\Leftrightarrow\left \{ {{x=0} \atop {g(x)=0}} \right.}$
Consider two cases:
*The first case: $g(0)=0\leftrightarrow m=\pm2$
For $m=2$, we have $f'(x)=8x^7$. The sign of $f'(x)$ changes from $-$ to $+$ at $x=0$, which means that $m=2$ satisfies. (1)
For $m=-2$, we have $f'(x)= 8 x^7-20 x^4=x^4(8x^3-20)$. This function doesn't change its sign at $x=0$, so $m=-2$ doesn't satisfy. (2)
*The second case: $g(x) \ne 0$
In order to reach local minimum at $x=0$, $f'(x)$ has to change its sign from $-$ to $+$ when $x$ passes through $0$.
This means that $g(x)$'s sign has to stay positive since $x^3$ changes its sign from $-$ to $+$ when $x$ passes through $0$, and $f(x)=x^3g(x)$
Which implies that: $\displaystyle{\left \{ {{\lim_{x \to 0^-}g(x)>0} \atop {\lim_{x \to 0^+}g(x)>0}} \right.}\Leftrightarrow 16-4m^2>0 \Leftrightarrow-2<m<2$ (3)
From (1), (2), and (3), we get: $-2<m \le 2$, $m \in \mathbb{Z}$, so $m \in \{-1,0,1,2\}$.
Is my solution correct? And is there any better way to this problem than this? Thanks!