I'm taking an introductory calculus course and was hoping to be able to prove the second derivative test using what I already know. I'm not sure if the steps I make are rational and wanted to verify.
Theorem statement
If $f'(x)=0$ and $f''(x)>0$ then $(x,f(x))$ is a local minimum of $f$.
If $f'(x)=0$ and $f''(x)<0$ then $(x,f(x))$ is a local minimum of $f$.
Step 1.1
If a function $f$ is continuous on the interval $[a,b]$ and $x\in (a,b)$ and $f(x)>b$ where $b$ is some constant real number, then for all $c$ near $x$, $f(c)>b$.
Assume this theorem this theorem were not true. Then the intermediate value theorem would not hold.
And similarly for the case of $f(x)<a$.
Step 1.2
Let the $f$ be continuous on $[a,b]$ and let $f''(x \in (a,b)) > 0$. Then by (1.1) it must be that for some open interval around $x$ containing $x$, $f''(x)>0$.
Step 1.3
If $f''(x)>0$ for some interval around $x$ containing x, and $f'(x)=0$, then for all $c$ in the interval, $c<x \implies f'(c)<0, c>x\implies f(c)>0$ because $f'(x)$ is an increasing function in this interval.
We now have that $x$ is a local minimum by the definition of a local minimum and can also consider the variant for $x$ being a local maximum.
I'm looking for two things:
- Help with the intervals. I don't have an intuitive grasp yet of which intervals to use. So I'd really appreciate if any redundant or incorrect intervals are used in my proof.
- Destroy the logic if I made any incorrect assumptions.
I know this is a very simple proof and shouldn't be that difficult. However as I'm still in the learning stage and my intuition hasn't matured, I'm not positive of my own abilities and want to know when I'm doing something wrong.
Thanks.
Best Answer
$0<f''(x)=\lim_{h\to 0^{+}}\frac{f'(x+h)-f'(x)}{h}=\lim_{h\to 0^{+}}\frac{f'(x+h)}{h}$ so there is a $\delta >0$ such that if $z\in (x,x+\delta )$ then $f'(z)>0$. Similarly, there is a $\delta_1>0$ such that if $z\in (x-\delta_1,x)$ then $f'(z)<0$.
Now apply the first derivative test.