Spelling Out Kathy’s Unit 6 Tamar ZEILBERGER

An alternative way to do Unit 6 #4 We have to prove that If x^{∗} is a local maximum of ln f(x) then it is also a local maximum of f(x). Putting g(x) = ln f(x), this is equivalent to the statement If x^{∗} is a 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 exponential function is an increasing function, e^{g(x)} has the same 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 chain rule

f′(x) = e^{g(x)} · g′(x)  (1)

In particular

f′(x^{∗}) = e^{g(x^{∗})} · g′(x^{∗}) = 0 .

So we know right away that x^{∗} is a 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 product rule to Eq. (1), we have

f′′(x) = (e^{g(x)} · g′(x))′ = (e^{g(x)})′ · g′(x) + e^{g(x)} · g′′(x)  (2)

Using Eq. (1) again we have

f′′(x) = e^{g(x)} · g′(x) · g′(x) + e^{g(x)} · g′′(x) = e^{g(x)} · g′(x)^{2} + e^{g(x)} · g′′(x)  (3)

Factoring out e^{g(x)} we finally get

f′′(x) = e^{g(x)}(g′(x)^{2} + g′′(x)) .  (3)

Plugging-in x = x^{∗} we get

f′′(x^{∗}) = e^{g(x^{∗})}(g′(x^{∗})^{2} + g′′(x^{∗})) .  (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^{∗})} · g′′(x^{∗}) .  (5)

Since e^{anything} is always positive, and by assumption g′′(x^{∗}) < 0, and since 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)}.

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

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

Proof: b − a is positive hence e^{b−a} > 1 hence e^{b}/e^{a} > 1 hence e^{b} > e^{a}