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I have been studying for the upcoming Introductory Real Analysis exam and got stuck upon the proof of the second derivative test. Here is a verbatim text of the theorem and the proof from the book:

THEOREM: Let $I$ be an open interval containing the point $x_0$ and suppose that the function $f:I\rightarrow R$ has a second derivative. Suppose that $ f'(x_0)=0$. If $f''(x_0)>0$, then $x_0$ is a local minimizer of $f:I \rightarrow R$.

PROOF: Since $f''(x_0)=\lim_{x\to{x_0}} \frac{f'(x)-f'(x_0)}{x-x_0}>0$, it follows (see Exercise 16) that there is a $\delta>0$ such that the open interval $(x_0-\delta, x_0+\delta)$ is contained in $I$ and $\frac{f'(x)-f'(x_0)}{x-x_0}>0$ if $0<|x-x_0|<\delta$. But $f'(x_0)=0$, so the (4.13) [preceding inequality] amounts to the assertion that if $|x-x_0|<\delta$, then $f'(x)>0 \text{ if } x>x_0$ and $f'(x)<0 \text{ if } x<x_0$. Using these two inequalities and the [Lagrange] Mean Value theorem, it follows that $f(x)>f(x_0) \text{ if } 0<|x-x_0|<\delta$.

The textbook states the Lagrange Mean Value Theorem as follows: Suppose the function $f:[a,b] \rightarrow R$ is continuous and $f: (a,b) \rightarrow R$ is differentiable. Then there is a point $x_0$ in the open interval $(a,b)$ at which $f'(x_0)=\frac{f(b)-f(a)}{b-a}$.

Luckily I was able to solve 'Exercise 16,' but I just don't see how I can apply the Lagrange Mean Value Theorem in the latter parts of the proof. Any small hints would be appreciated. Thanks in advanced.

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    $\begingroup$ Looking at the Cauchy Mean Value Theorem, it seems to be the same as the Lagrange MVT. In any case, try stating that theorem in the question and see if you can tell how it might apply. $\endgroup$ – robjohn May 6 '19 at 2:34
  • $\begingroup$ Sure, hang on. Thanks for the heads up. $\endgroup$ – Ko Byeongmin May 6 '19 at 2:34
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First, apply your MVT to the interval $I=(x_0,x_0+\delta)$. Take an $x\in I$. The MVT says that there is a $c, x_0<c<x$ such that: $f'(c)=\frac{f(x)-f(x_0)}{x-x_0} $.

Since $f'(c) >0$ (it's one of your inequalities), then $f(x)>f(x_0)$. A similar argument holds for the interval $I'=(x_0-\delta,x_0)$.

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If $x_0<x<x_0+\delta$, apply the MVT with $a=x_0$, $b=x$. You get $f(x)-f(x_0)=(x-x_0)f'(y_0)$ for some $y_0\in (x_0,x)$. Since the assertion tells you that $f'(y_0)>0$, you have $f(x)-f(x_0)>0$.

Do the same thing in the other case $x_0-\delta<x<x_0$.

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