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!