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The following is a problem from chapter 11, "Significance of the Derivative", from Spivak's Calculus

  1. Redo problem 10-18(c) when

$$(f')^2=f-\frac{1}{f^2}\tag{1}$$

Here is problem 10-18(c)

10-18. (c) Suppose that the function $f>0$ has the property that

$$(f')^2=f+\frac{1}{f^3}\tag{2}$$

Find a formula for $f''$ in terms of $f$.

10-18 is relatively straightforward to solve.

First, note that if $f'(x)=f(x)+\frac{1}{f(x)^3}=0$ then $f(x)^4=-1$, which never occurs for any $x$. Hence $f' \neq 0$.

Just take the derivative of both sides of $(2)$

$$2f'f''=f'-3f^{-4}f'$$

Therefore $f''(x)=\frac{1}{2}-\frac{3}{2f^{(4)}(x)}$.

Let's consider the solution to 59.

First, note that if $f'(x)=0$ then $0=f(x)-\frac{1}{f^2(x)} \implies f(x)=1$

Differentiate both sides of $(1)$

$2f'(x)f''(x)=f'(x)+\frac{2f'(x)}{f^3(x)}$

Dividing by $f'(x)$, we have for $f(x) \neq 1$ that

$$f(x)''=\frac{1}{2}+\frac{1}{f(x)^3}\tag{3}$$

At this point, according to the solution manual we can show that $(3)$ also holds if $f(x) = 1$, by applying the following theorem

Theorem 7: Suppose $f$ is continuous at $a$, $f'(x)$ exists for all $x$ in some interval containing $a$, except perhaps for $x=a$. Suppose, moreover, that $\lim\limits_{x \to a} f'(x)$ exists. Then $f'(a)$ also exists, and

$$f'(a)=\lim\limits_{x \to a} f'(x)$$

How do we show that $(3)$ also holds for $f(x)=1$ using this theorem?

xoux
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1 Answers1

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I do not think that Theorem 7 suffices to prove $(3)$. As you know, we have a problem for all points $a$ with $f'(a) = 0$ (which is equivalent to $f(a) = 1$). We can try to apply Theorem 7 to $\phi(x) = f'(x)$. If $\phi'(x) = f''(x)$ exists for all $x$ in some interval containing $a$, except perhaps for $x=a$, then $\phi'(a) = f''(a)$ also exists and $$f''(a) = \lim_{x \to a} f''(x) = \lim_{x \to a}\left( \frac 1 2 + \frac{1}{f(x)^3} \right) = \frac 1 2 + \frac{1}{f(a)^3} .$$ But when does $f''(x)$ exist in some interval containing $a$, except perhaps for $x=a$? Our present knowledge only covers the case that $f'(x) \ne 0$ ($\Leftrightarrow f(x) \ne 1$) in some interval containing $a$, except for $x = a$.

Thus if the set $A = f^{-1}(1)$ only has isolated points, we are done. But if $A$ has non-isolated points, we need new arguments. These can be found in

"Calculus 4th Edition" by Michael Spivak -- Chapter 11 Problem 59
Spivak — Chapter 11 Problem 59 Problems understanding the solution

Paul Frost
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