Example 28: Absolute extrema of $12x^{4/3}-6x^{1/3}$ on $[-1,1]$

Basics you need

Solution

1. What the function means, and why the answer exists

   For every real $x$⁠, $x^{1/3}$ is the real cube root, so $f$ is defined on all of $[-1, 1]$⁠. Writing $12x^{4/3} = 12x\cdot x^{1/3}$⁠,

   $$f(x) = 6x^{\frac13}(2x - 1).$$

   $f$ is built from continuous functions, so it is continuous on $[-1, 1]$⁠, and by Theorem 5 it has an absolute maximum and an absolute minimum there.

   Negative values of $x$ are allowed, so it is important to read $x^{1/3}$ as the real cube root ($(-1)^{1/3} = -1$⁠). The factorised form helps later: it shows $f(0) = 0$ and $f\left(\tfrac12\right) = 0$ at a glance.
   $x^{4/3} = x^{1 + 1/3} = x \cdot x^{1/3}$ (laws of indices, Basic 1).

2. The derivative for $x \ne 0$

   By the power rule, for $x \ne 0$⁠:

   $$\begin{aligned} f'(x) &= 12\cdot\tfrac43\,x^{\frac13} - 6\cdot\tfrac13\,x^{-\frac23} \\ &= 16x^{\frac13} - \frac{2}{x^{2/3}} \\ &= \frac{16x - 2}{x^{2/3}} = \frac{2(8x - 1)}{x^{2/3}}. \end{aligned}$$
   To combine the two terms we multiply the first by $\frac{x^{2/3}}{x^{2/3}}$⁠: $16x^{1/3}\cdot x^{2/3} = 16x$⁠. The single fraction shows where $f' = 0$ (numerator zero) and where the formula breaks down (denominator zero).
   Power rule (Basic 2): $\frac{d}{dx}x^{4/3} = \frac43 x^{1/3}$⁠, $\frac{d}{dx}x^{1/3} = \frac13 x^{-2/3}$⁠.

3. $f$ is not differentiable at $x = 0$

   The formula for $f'$ has $x^{2/3}$ in the denominator, so it says nothing at 0. Use the definition. For $h \ne 0$⁠,

   $$\frac{f(h) - f(0)}{h} = \frac{12h^{4/3} - 6h^{1/3}}{h} = 12h^{\frac13} - \frac{6}{h^{2/3}}.$$

   As $h \to 0$⁠, $12h^{1/3} \to 0$ while $\frac{6}{h^{2/3}} \to +\infty$ (from both sides, since $h^{2/3} > 0$⁠). So the difference quotient tends to $-\infty$⁠: $f'(0)$ does not exist, and the graph has a vertical tangent at the origin.

   The book simply notes that $f'(x)$ is not defined at $x = 0$⁠. The difference quotient makes this precise: an infinite slope is not a derivative. Because $f$ is not differentiable at 0, Theorem 6 does not apply there, and 0 must be on the list of candidates.
   Derivative as a limit of the difference quotient (Chapter 5). Basics: rate of change as a derivative

4. All the critical points (rule step 1)

   For $x \ne 0$⁠, $f'(x) = 0$ exactly when $8x - 1 = 0$⁠, that is $x = \frac18$⁠. At $x = 0$⁠, $f$ is not differentiable. So the critical points in $[-1, 1]$ are

   $$x = 0 \quad\text{and}\quad x = \tfrac18.$$
   Both kinds of critical point count. A fraction is zero only where its numerator is zero and its denominator is not; the denominator $x^{2/3}$ is not zero at $\frac18$⁠.
   Critical point: $f'(c) = 0$ or $f$ not differentiable at $c$ (Note after Theorem 2, Section 6.4). Lesson: maxima and minima

5. Values at the critical points and end points (rule steps 2 and 3)

   The end points are $-1$ and $1$⁠. Using $(-1)^{1/3} = -1$⁠, $\left(\tfrac18\right)^{1/3} = \tfrac12$⁠:

   $$\begin{aligned} f(-1) &= 12(-1)^4 - 6(-1) = 12 + 6 = 18, \\ f(0) &= 0 - 0 = 0, \\ f\!\left(\tfrac18\right) &= 12\cdot\tfrac{1}{16} - 6\cdot\tfrac12 = -\tfrac94, \\ f(1) &= 12 - 6 = 6. \end{aligned}$$
   Compute the cube root first and then raise it to the 4th power: $\left(\tfrac18\right)^{4/3} = \left(\tfrac12\right)^4 = \tfrac1{16}$⁠, and $(-1)^{4/3} = (-1)^4 = 1$⁠. Then $\frac{12}{16} - 3 = \frac34 - 3 = -\frac94$⁠.
   $x^{m/3} = \left(x^{1/3}\right)^m$ (Basic 1).

6. Pick the largest and smallest (rule step 4)

   Of $18,\ 0,\ -\frac94,\ 6$⁠, the largest is 18 and the smallest is $-\frac94$⁠. So the absolute maximum value is 18, at $x = -1$⁠, and the absolute minimum value is $-\frac94$⁠, at $x = \frac18$.

   The sign of $f'(x) = \frac{2(8x - 1)}{x^{2/3}}$ confirms the picture: the denominator is positive for $x \ne 0$⁠, so $f' < 0$ on $[-1, 0)$ and $(0, \frac18)$ and $f' > 0$ on $(\frac18, 1]$⁠. So $f$ falls from 18 all the way down to $-\frac94$ and then rises to 6. The point $x = 0$ is a critical point but not an extremum: $f$ keeps falling through it.
   Working Rule, Section 6.4.1. Lesson: absolute extrema on a closed interval

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