Find points of relative extrema, the intervals of increase & decrease for $k(x)=x^4+2x^2-4$ This is a real analysis problem, so I want to know how to make my solution rigorous in the appropriate way. Find points of relative extrema, the intervals on which the function is increasing & decreasing on for $k(x)=x^4+2x^2-4$, $f\colon\mathbb{R}\to\mathbb{R}$. 

The First Derivative Test for Extrema states: Let $f$ be continuous on the interval $I=[a,b]$ and let $c$ be an interior point of $I$. Assume that $f$ is differentiable on $(a,c)$ and $(c,b)$. Then:
  
  
*
  
*If there is a neighborhood $(c-d,c+d)\subseteq I$ such that $f '(x)\geq 0$ for $c-d < x < c$ and $f'(x)\leq 0$ for $c < x < c+d$, then $f$ has a relative maximum at $c$.
  
*If there is a neighborhood $(c-d,c+d)\subseteq I$ such that $f '(x)\leq 0$ for$ c-d < x < c$ and $f'(x)\geq 0$ for $c < x < c+d$, then $f$ has a relative minimum at $c$.

Doing it the old fashion Calculus way first I obtain,
$$\begin{align*}
k(x)&=x^4+2x^2-4\\
k'(x)&=4x^3+4x
\end{align*}$$
Find the critical points:
$k'(x)=4x(x^2+1)$. Set:
$$\begin{align*}
4x = 0 &\implies x=0\\
x^2+1 = 0 &\implies x\text{ is not a real number}
\end{align*}$$
So, the critical point is $x=0$.
To test whether $0$ is a relative max/min,
I check numbers to the left/right of it so:


*

*$f'(-1)=-8$ 

*$f'(1)=8$


So, this meets condition 2. $f$ has a local minimum at $x=0$.
The function is decreasing on $(-\infty,0)$ & increasing on $(0,\infty)$
Given the old calculus way to do it, how do I prove this result using the condition :


*

*If there is a neighborhood $(c-d,c+d) \subseteq I$ such that $f'(x)\leq 0$ for $c-d < x < c$ and $f'(x)\geq 0$ for $c < x < c+d$, then $f$ has a relative minimum at $c$.


How do I pick my $c,d$ and complete the proof? I am at lost on this, but I feel like it should be fairly simple. Thanks!
 A: Essentially, what you do with "old calculus" is doing this: notice that $f'(x)$ is continuous. By the Intermediate Value Theorem, it can only change signs if it goes through $0$; since the only place it is equal to $0$ is at $0$, it always has the same sign on $(-\infty,0)$, and it always has the same sign on $(0,\infty)$. Evaluating at $-1$ tells you the sign on $(-\infty,0)$, evaluating at $1$ tells you the sign on $(0,\infty)$.
You can now verify that if you pick $c=0$ and $d=1$ (or $d$ any positive value), you will satisfy the condition. The point $c$ must be the critical point; the value $d$ is just any positive value small enough that $(c-d,c+d)$ does not get you "out of" the interval $I$, and where the derivative does not change signs except at $c$. Here, your derivative only changes signs in one place and your interval is the entire real line, so both of these "requirements" become vacuous: any $d\gt 0$ will work.
A: Note that $k(x)=(x^2+1)^2-5=f(x^2+1)$ with $f(u)=u^2-5$. The function $f$ is increasing on $u\gt0$ and $x\mapsto x^2+1$ has positive values, is decreasing on $x\leqslant0$ and increasing on $x\geqslant0$. This proves that $x\mapsto k(x)$ is decreasing on $x\leqslant0$ and increasing on $x\geqslant0$ and that its only local extremum is a global infimum, located at $x=0$.
