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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:

  1. 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 $c$.

  2. 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:

  1. f'(-1)=-8
  2. 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 :

  1. 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!

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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.

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    @Quaternary: But we know that the latter does not happen, since $c\neq 0\implies f'(c)\neq 0$. Therefore, given any two points $a$ and $b$, $0\lt a\lt b$, we have that $f'(a)$ and $f'(b)$ are either both positive, or both negative. Which one is it? Since $1\gt 0$, if $b\gt 0$ then $f'(1)$ and $f'(b)$ have the same sign (both positive, or both negative). And since $f'(1)\gt 0$, we conclude that if $b\gt 0$, then $f'(b)\gt 0$. A similar argument shows that if $b\lt 0$, then $f'(b)$ and $f'(-1)$ must have the same sign (both positive or both negative), and since $f'(-1)\lt 0$...2012-04-07
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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 $x=0$.