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\centerline
{\bf  Spelling Out Kathy's Unit 6 }
\bigskip
\centerline
{\it Tamar ZEILBERGER}


{\bf An alternative way to do  Unit 6 \#4}

We have to prove that 

If $x^{*}$ is a {\bf local maximum} of $\ln f(x)$ then it is also a local maximum of $f(x)$.

Putting $g(x)=\ln f(x)$, this is {\bf equivalent} to the statement

If $x^{*}$ is a {\bf local maximum} of $g(x)$ then it is also a local maximum of $e^{g(x)}$.

(This is true because, of course, $f(x)=e^{\ln f(x)}$.)

But you don't need calculus for that! Pre-calculus suffices. Since the {\bf exponential} function is
an {\bf increasing} function, $e^{g(x)}$ has the same {\bf ups and downs} as $g(x)$.

So it is clear that $g(x)$ and $e^{g(x)}$ share their maxima (and minima!).

But, if you want to use calculus, you can.

We are given that $g'(x^{*})=0$, $g''(x^{*})<0$.

Since $f(x)=e^{g(x)}$, we have, by the {\bf chain rule}

$$
f'(x)= e^{g(x)} \cdot g'(x)
\eqno(1)
$$
 In particular
$$
f'(x^{*})= e^{g(x^{*})} \cdot g'(x^{*}) =0 \quad.
$$
So we know right away that $x^{*}$ is a {\bf critical point} of $f(x)$.

To see whether it is a  max or min, we need to express $f''(x)$ in terms of $g(x)$ and its derivatives.

Applying the {\bf product rule} to Eq. $(1)$, we have

$$
f''(x) \, = \, ( e^{g(x)} \cdot g'(x) )' \, = \, (e^{g(x)})' \cdot g'(x) \, + \,  e^{g(x)} \cdot g''(x)
\eqno(2)
$$

Using Eq. (1) again we have

$$
f''(x) \, = \,  e^{g(x)} \cdot g'(x) \cdot g'(x) \, + \,  e^{g(x)} \cdot g''(x)
= \,  e^{g(x)} \cdot g'(x)^2 \, + \,  e^{g(x)} \cdot g''(x)
\eqno(3)
$$
Factoring out  $e^{g(x)}$ we finally get
$$
f''(x)\, = \,  e^{g(x)} ( g'(x)^2 \, +  g''(x)) \quad .
\eqno(3)
$$
Plugging-in $x=x^{*}$ we get
$$
f''(x^{*})\, = \,  e^{g(x^{*})} ( g'(x^{*})^2 \, +  g''(x^{*})) \quad .
\eqno(4)
$$
But, we already know that  $g'(x^{*})=0$, so
$$
f''(x^{*})\, = \,  e^{g(x^{*})} ( 0^2 \, +  g''(x^{*})) 
\, = \,  e^{g(x^{*})} ( 0 \, +  g''(x^{*})) \, = \,  e^{g(x^{*})}  \cdot g''(x^{*}) 
\quad .
\eqno(5)
$$
Since $e^{anything}$ is always {\bf positive}, and by assumption  $g''(x^{*}) <0$, and since {\it positive times negative is negative} ,
we proved that $f''(x^{*})<0$. Combined with the above fact that $f'(x^{*})=0$, this proves that $x^{*}$ is also
a local maximum of $f(x)=e^{g(x)}$.

{\bf Comment}: No offense to calculus, the above proof using {\bf precalculus} is much better and
more insightful. To formally prove that the exponential function $e^x$ is an {\bf increasing function}
you could of course take the derivative $(e^x)'=e^x$ and argue that it is always positive,
but using {\bf high school algebra} it is obvious that

{\bf Precalculus Lemma}: If $b>a$ then $e^b>e^a$.

{\bf Proof}: $b-a$ is positive hence $e^{b-a}>1$ hence $e^b/e^a>1$ hence $e^b>e^a$

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