Miscellaneous · Q18: $f(x)=|x|^3$ has $f''(x)=6|x|$ for all real $x$

Basics you need

Solution

1. Write $f$ without the absolute value

   $$f(x) = |x|^3 = \begin{cases} x^3, & x \ge 0, \\ -x^3, & x < 0. \end{cases}$$
   For $x < 0$⁠, $|x| = -x$⁠, so $|x|^3 = (-x)^3 = -x^3$⁠. Note that $-x^3$ is positive when $x$ is negative, as $|x|^3$ must be.
   $|x|$ piecewise (Basic 1). Basics: piecewise functions

2. $f'(x)$ for $x \ne 0$

   For $x > 0$⁠: $f'(x) = 3x^2$⁠. For $x < 0$⁠: $f'(x) = -3x^2$⁠.

   Take $x > 0$⁠. For all small enough $h$ ($|h| < x$⁠), both $x$ and $x + h$ are positive, so the difference quotient $\frac{f(x+h) - f(x)}{h}$ only ever uses the formula $x^3$⁠. Its limit is therefore the derivative of $x^3$⁠. The same argument works for $x < 0$ with $-x^3$⁠. At $x = 0$ this fails, because $h$ of either sign takes us onto different formulas; that is why 0 is treated separately.
   Power rule (Basic 3) and constant multiple rule, Section 5.3. Basics: derivative rules

3. $f'(0)$ from the two one-sided difference quotients

   Here $f(0) = 0$⁠, so the difference quotient is $\frac{f(h) - 0}{h} = \frac{|h|^3}{h}$⁠.

   Left ($h < 0$⁠): $|h|^3 = -h^3$⁠, so

   $$\lim_{h\to0^-}\frac{-h^3}{h} = \lim_{h\to0^-}(-h^2) = 0.$$

   Right ($h > 0$⁠): $|h|^3 = h^3$⁠, so

   $$\lim_{h\to0^+}\frac{h^3}{h} = \lim_{h\to0^+}h^2 = 0.$$

   Both are finite and equal, so $f$ is differentiable at 0 and $f'(0) = 0$⁠.

   This is the book's definition of differentiability at a point, used exactly as in its discussion of $|x|$ after Theorem 3. For $|x|$ itself the two quotients were $-1$ and $1$ (a corner). For $|x|^3$ both are 0: cubing flattens the corner completely.
   Left-hand and right-hand derivatives, Section 5.3 (Basic 2). Lesson: differentiability

4. One formula for $f'$

   $$f'(x) = \begin{cases} 3x^2, & x > 0, \\ 0, & x = 0, \\ -3x^2, & x < 0, \end{cases}$$

   that is, in one formula,

   $$f'(x) = 3x|x| \quad \text{for every real } x.$$
   Check each piece: for $x > 0$⁠, $3x|x| = 3x^2$⁠; for $x < 0$⁠, $3x|x| = 3x(-x) = -3x^2$⁠; at 0 it gives 0. So $f'$ exists at every real $x$⁠. The compact form $3x|x|$ is handy for the next derivative.
   $|x|$ piecewise (Basic 1).

5. $f''(x)$ for $x \ne 0$

   For $x > 0$⁠: $f''(x) = \frac{d}{dx}(3x^2) = 6x$⁠. For $x < 0$⁠: $f''(x) = \frac{d}{dx}(-3x^2) = -6x$⁠.

   The same reasoning as Step 2: on the open interval $x > 0$⁠, $f'$ agrees with the polynomial $3x^2$⁠, so its derivative there is that polynomial's derivative; likewise for $x < 0$⁠.
   Power rule (Table 5.3); second derivative is the derivative of $f'$ (Section 5.7). Lesson: second order derivative

6. $f''(0)$ from the one-sided difference quotients of $f'$

   Now apply the definition to $f'$ at 0. Since $f'(0) = 0$⁠, the quotient is $\frac{f'(h) - 0}{h} = \frac{3h|h|}{h} = 3|h|$ for $h \ne 0$⁠.

   Left ($h < 0$⁠):

   $$\lim_{h\to0^-}3|h| = \lim_{h\to0^-}(-3h) = 0.$$

   Right ($h > 0$⁠):

   $$\lim_{h\to0^+}3|h| = \lim_{h\to0^+}3h = 0.$$

   Both are 0, so $f'$ is differentiable at 0 and $f''(0) = 0$⁠.

   This is the heart of the question. We cannot get $f''(0)$ by plugging $x = 0$ into "$6x$" or "$-6x$": those formulas were derived only for $x \ne 0$⁠. The definition must be applied to $f'$ at the junction, and it confirms that both sides agree.
   Definition of the derivative at a point, applied to $f'$ (Basics 2 and 4). Basics: left-hand and right-hand limits

7. Conclude: $f''(x) = 6|x|$ for every real $x$

   $$f''(x) = \begin{cases} 6x, & x > 0, \\ 0, & x = 0, \\ -6x, & x < 0, \end{cases}$$

   that is, in one formula,

   $$f''(x) = 6|x| \quad \text{for every real } x.$$
   $f''$ exists at every real number: Step 5 handles $x \ne 0$ and Step 6 handles $x = 0$⁠. The single formula $6|x|$ matches all three pieces. Going one step further, $f'' = 6|x|$ has a corner at 0 (its one-sided difference quotients are $-6$ and $6$⁠), so the third derivative of $|x|^3$ does not exist at 0. Each differentiation costs one order of smoothness: $f$ can be differentiated twice at 0, $f'$ once, $f''$ not at all.
   Same corner as $|x|$ at 0 (discussion after Theorem 3, Section 5.3). Lesson: differentiability

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