In Michael Spivak's "Calculus" 4th edition, Chapter 11 Problem 59 reads
Suppose that the function $f > 0$ has the property that $$(f')^2=f-{1 \over f^2}.$$ Find a formula for $f''$ in terms of $f$. (Why is this problem in this chapter?)
This seems to be straight-forward:
\begin{align}
(f')^2 &= f-{1 \over f^2} \\
2f'f'' &= f'+{2 \over f^3}f' \\
2f'' &= 1+{2 \over f^3} \\
f'' &= {1 \over 2} + {1 \over f^3}
\end{align}
My problem is I can't figure out how to answer why this problem is in this chapter. Chapter 11 covers Rolle's Theorem, the Mean Value Theorem, the Cauchy Mean Value Theorem and L'Hopital's Rule. It also covers local extrema and the Second-Derivative Test. We answered a substantially similar question in Chapter 10 after covering the Chain Rule etc. Since I'm not using any new concepts, I'm suspicious I'm doing something wrong.
I'd very much appreciate any insight.
EDIT: Thank you everyone, you have all been very helpful. I would accept both answers if I could.
Best Answer
You do not say where $f$ is defined, so let us assume it is defined on an open (bounded or unbounded) interval $D = (a,b)$. You can also take a half-open or a closed interval if you are willing to work with left-hand / right-hand derivatives in the boundary points $a$ or $b$.
Let us introduce the continuos function $$g = \frac{1}{2} +\frac{1}{f^3} .$$ Note that $g > \frac{1}{2}$.
You claimed that $f'' = g$, but the proof breaks down in the points $\xi \in D$ where $f'(\xi) = 0$. However we shall prove the following:
Either $f = 1$ (in which case $f''(\xi) = 0 \ne g(\xi)$ for all $\xi$) or $f \ne 1$ and $f'' = g$.
We have $f'(x)^2 \ge 0$, hence $f(x) - \frac{1}{f(x)^2} \ge 0$ which means $f(x)^3 \ge 1$, that is, $f \ge 1$. Let $A = f^{-1}(1)$. This is a closed subset of $D$. We have $\xi \in A$ iff $f'(\xi) = 0$. Thus for all $\xi \in D \setminus A$ we correctly get $$(*) \quad f''(\xi) = \frac{1}{2} +\frac{1}{f(\xi)^3} = g(\xi) .$$ What happens if $\xi \in A$?
Case 1. $\xi$ is not a limit point of $A$, that is, $\xi$ is an isolated point of $A$.
Then a pointed open neigborhood $U = (\xi - \varepsilon, \xi + \varepsilon) \setminus \{ \xi \}$ of $\xi$ is contained in $D \setminus A$. Now we apply Rolle's theorem. For $x \in U$ we have $$ \frac{f'(x) - f'(\xi)}{x - \xi} = f''(\zeta(x)) = g(\zeta(x))$$ for some $\zeta(x)$ between $x$ and $\xi$. If $x \to \xi$, then clearly $\zeta(x) \to \xi$. Hence $$f''(\xi) = \lim_{x \to \xi} \frac{f'(x) - f'(\xi)}{x - \xi} = \lim_{\zeta(x) \to \xi} g(\zeta(x)) = g(\xi) .$$ Thus $(*)$ is also correct for $\xi \in A$ being an isolated point of $A$.
Case 2. $\xi$ is a limit point of $A$, that is, $\xi \in A' =$ set of limit points of $A$.
Then we find a sequence $(x_n)$ in $A \setminus \{ \xi \}$ such that $x_n \to \xi$. Therefore $$f''(\xi) = \lim_{n\to \infty}\frac{f'(x_n) - f'(\xi)}{x_n - \xi} = 0$$ because $f'(x_n) = f'(\xi) = 0$.
Summarizing we have $f''(\xi) = 0$ iff $\xi \in A'$ and $f''(\xi) = g(\xi) > \frac{1}{2}$ iff $\xi \notin A'$.
We claim that if $A' \ne \emptyset$, then $A = D$ which means $f = 1$. This is the alternative solution of Mohammad Riazi-Kermani's answer.
So pick any $\xi \in A'$ and let $C$ be the connected component of $\xi$ in $A$. Since $A$ is closed in $D$, also $C$ is closed in $D = (a,b)$. As a connected subset of the real line $C$ must be an interval with boundary points $c \le d$.
Assume $a < c$. Then $c \in C$ because $C$ is closed in $D$. Moreover $c \in A'$. This is clear for $c < d$ and also for $c = d$ (in this case we have $\xi \in [c,c]$, thus $c = \xi \in A'$). For each $0 < \varepsilon < c-a$ the interval $(c-\varepsilon,c)$ must contain a point of $D \setminus A$ (otherwise $C' = (c-\varepsilon,c) \cup C$ would be a connected subset of $A$, hence $C' \subset C$ which is a contradiction). Thus we can find a sequence $(x_n)$ in $D \setminus A$ such that $x_n < c$ and $x_n \to c$. Then again by Rolle's theorem $$\frac{f'(x_n) - f'(c)}{x - c} = f''(\zeta(x_n))$$ for some $\zeta(x_n)$ between $x_n$ and $c$. Since $f'(x_n) - f'(c) = f'(x_n) \ne 0$, we see that $f''(\zeta(x_n)) \ne 0$ and conclude $\zeta(x_n) \notin A'$. Hence $$\frac{f'(x_n) - f'(c)}{x - c} = g(\zeta(x_n) .$$ We have $\zeta(x_n) \to c$ as $n \to \infty$ so that $$f''(c) = \lim_{n \to \infty} \frac{f'(x_n) - f'(c)}{x_n - c} = \lim_{n \to \infty} g(\zeta(x_n)) = g(c) \ne 0 .$$ But on the other hand $f''(c) = 0$ because $c \in A'$. This is a contradiction.
Therefore $a = c$.
Similarly we show that $b = d$. Hence $C = D$ which shows $A = D$ and $f = 1$.
If $f \ne 1$, then $A \ne D$. We have shown that if $A' \ne \emptyset$, then $A = D$, i.e. $f = 1$, which is impossible. Hence $A' = \emptyset$ and $f'' = g$.