Q.Find absolute maximum and minimum values of a function $f$ given by $f(x) = 12x^{4/3} - 6x^{1/3},\ x \in [-1, 1]$.

Idea: on a closed interval, compare $f$ at its critical points (where $f'=0$ or $f'$ is undefined) and at the endpoints.

Let $f(x)=12x^{4/3}-6x^{1/3}$ on $[-1,1]$.

Step 1 — Derivative.

Step 2 — Critical points. $f'(x)=0$ gives $8x-1=0\Rightarrow x=\tfrac18$; and $f'$ is undefined at $x=0$. Both lie in $[-1,1]$.

Step 3 — Evaluate at $x=-1,0,\tfrac18,1$.

$f(-1)=12(1)-6(-1)=18,\qquad f(0)=0,$ …

The plan

The cube root $x^{1/3}$ — and hence $x^{4/3}$ — is defined and continuous for every real number, so $f$ is continuous on the closed interval $[-1,1]$ and must attain an absolute maximum and minimum there. These occur only at a critical point (where $f'(x)=0$ or where $f'(x)$ fails to exist) or at an endpoint. We list all such $x$, evaluate $f$, and compare.

Step 1 — Differentiate

Factor out the lower power $2x^{-2/3}$:

Step 2 — Critical points

• $f'(x)=0$: since $x^{-2/3}\ne 0$ for $x\ne0$, we need $8x-1=0$, i.e. $x=\tfrac18$.
• $f'$ undefined: $x^{-2/3}$ blows up at $x=0$, so $x=0$ is also a critical point (the graph has a cusp there, but $f$ itself is still defined and continuous).

Both $x=0$ and $x=\tfrac18$ lie in $[-1,1]$.

Step 3 — Evaluate $f$ at all candidates

Candidates: endpoints $x=-1,1$ and critical points $x=0,\tfrac18$.

• At $x=-1$: $(-1)^{1/3}=-1$ and $(-1)^{4/3}=\big((-1)^{1/3}\big)^4=(-1)^4=1$, so

• At $x=0$: $f(0)=0-0=0.$ …

Method: Absolute Extrema with Fractional-Power Derivatives

The same underlying technique as testing critical points and endpoints on a closed interval, but this method highlights the extra care needed when a function involves fractional exponents like $x^{1/3}$ or $x^{-2/3}$.

Steps

Step 1: Differentiate using the power rule for fractional exponents

Apply $\frac{d}{dx}x^n = nx^{n-1}$ exactly as usual, even when $n$ is a fraction. This often produces a negative power of $x$ in the result, which is the source of the extra care needed in the next step.

Step 2: Factor out the lowest power to reveal both kinds of critical point

Factoring this way makes two things visible at once: where the bracket vanishes ($f'(x)=0$), and where the factored-out power itself blows up ($f'(x)$ undefined). Both are genuine critical points — a critical point is any $x$ in the domain where $f'(x)=0$ or $f'(x)$ fails to exist, not only where it equals zero.

Step 3: List every candidate honestly …

Common Mistakes

Mistake 1: Only solving $f'(x)=0$ and missing the point where $f'(x)$ is undefined

Why it's wrong: for $f(x)=12x^{4/3}-6x^{1/3}$, the derivative $f'(x)=2x^{-2/3}(8x-1)$ has a negative power of $x$, which is undefined at $x=0$ even though $f(0)$ itself is perfectly defined ($f(0)=0$). A critical point is any $x$ where $f'(x)=0$ or $f'(x)$ does not exist — skipping $x=0$ would miss a real candidate entirely. Correct approach: after differentiating, explicitly check the domain of $f'(x)$ for any values that make it undefined, not just the zeros of the numerator.

Mistake 2: Sign errors when evaluating odd roots of negative numbers …