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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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+1 for showing the work. –  Arturo Magidin Apr 7 '12 at 15:52
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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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I have a few questions. I know it is obvious that f'(x) is cts, but what justification in my proof do I need to say this? Also, within my proof should I do as I did above (find f'(x) & then the critical pts) & mention 'By IVT f'(x) changes signs if it goes through 0, since f'(-1)<=0 then f'(x) decreases on (-inf,0) and since f'(1)>=0 then f'(x) increases on (0,inf)? For verifying the condition does it suffice to say there is a d-neighborhood st (0-2,0+2) subset I st f(-1)<=0 for -2<-1<0 and f'(1)>=0 st 0 < 1 <1+2 then f has a relative min at 0? Here c=0, d=2 –  Quaternary Apr 7 '12 at 16:11
    
@Quaternary: What justification you need depends on what you know (but note that you don't need to do it to apply the First Derivative Test as you quote it; but it can help you figure out why any $d\gt 0$ will work in this case; you do need it to argue that $f(x)$ is decreasing on $(-\infty,0)$. If you know polynomials are continuous everywhere, you can invoke that (but again, you don't have to in order to apply the 1DT). Also: it is false that $f'(x)$ decreases on $(-\infty,0)$: it is $f(x)$ that is decreasing on that interval. (cont) –  Arturo Magidin Apr 7 '12 at 16:14
    
@Quaternary: No, for verifying the condition as written, you need to show that for any $a$ on $(c-d,c) = (-2,0)$ you have $f'(a)\leq 0$; and that for any $b$ on $(c,c+d) = (0,2)$ you have $f'(b)\geq 0$. Just evaluating at $-1$ and at $1$ doesn't prove it unless you explicitly use the IVT to explain why this means that $f'(x)$ is never positive on $(-2,0)$ and never negative on $(0,2)$. –  Arturo Magidin Apr 7 '12 at 16:16
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@Quaternary: Suppose there exist points $a$ and $b$ with $f'(a)\gt 0$ and $f'(b)\lt 0$; then there has to be a point $r$ between $a$ and $b$ where $f'(b)=0$ (since $f'(x)$ is continuous everywhere). Since $f'(x)\neq 0$ for all $x$ on $(-\infty,0)$, that means that either $f'(x)\gt 0$ for all $x$ on $(-\infty,0)$, or $f'(x)\lt 0$ for all $x$ on $(-\infty,0)$. By noting that $f(-1)\lt 0$, we determine which of the two mutually exclusive possibilities we are in. –  Arturo Magidin Apr 7 '12 at 16:31
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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 $f'(-1)\lt 0$... –  Arturo Magidin Apr 7 '12 at 20:58
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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 $x=0$.

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