Q.Find the value of the following: The maximum value of $[x(x-1)+1]^{\frac{1}{3}}$, $0 \le x \le 1$ is (A) $(\frac{1}{3})^{\frac{1}{3}}$ (B) $\frac{1}{2}$ (C) $1$ (D) $0$ Miscellaneous Examples

Concept: Quadratic Extrema – the cubic root is monotonic, so the maximum of the whole expression occurs where the quadratic inside is maximum.

1. Let $f(x) = x(x-1)+1 = x^2 - x + 1$. This is a parabola opening upward.
2. On $[0,1]$, the vertex is at $x = \frac{1}{2}$. Since the parabola opens upward, the maximum on a closed interval occurs at an endpoint. …

The expression is $f(x) = \bigl[x(x-1)+1\bigr]^{1/3}$. Since the cube root function $t \mapsto t^{1/3}$ is strictly increasing for all real $t$, the value of $f(x)$ is largest exactly when the quantity inside the brackets is largest. So the problem reduces to a much simpler one: find the maximum of the quadratic $g(x) = x(x-1)+1$ on the closed interval $0 \le x \le 1$, then take its cube root.

Let’s rewrite $g(x)$ in standard form:

This is a parabola opening upward (coefficient of $x^2$ is positive). For an upward-opening parabola, the vertex gives the minimum, not the maximum. On a closed interval, the maximum of such a function occurs at one of the endpoints.

1. Find the vertex (just to confirm it’s a minimum):

   The vertex is at $x = -\frac{b}{2a} = -\frac{(-1)}{2(1)} = \frac{1}{2}$.

   At $x = \frac12$, $g\left(\frac12\right) = \frac14 - \frac12 + 1 = \frac34$.

   So the minimum value of $g(x)$ on $\mathbb{R}$ is $\frac34$, which is inside our interval.

2. Evaluate at the endpoints:

   At $x = 0$: $g(0) = 0 - 0 + 1 = 1$.

   At $x = 1$: $g(1) = 1 - 1 + 1 = 1$.

   Both endpoints give $g(x) = 1$.

3. Compare with the interior: …

Method: Optimizing a Monotonic Function of a Simpler Inner Expression

When the quantity to optimize is written as (something simple) raised to a power, or passed through any function that is strictly increasing, you do not need to differentiate the whole complicated expression — you only need to optimize the simpler inner expression.

Steps

Step 1: Identify the outer function and check it is monotonic increasing

Here the outer function is the cube root, $t\mapsto t^{1/3}$, which is strictly increasing for all real $t$. Because it never decreases, the largest output always comes from the largest input.

Step 2: Reduce the problem to optimizing the inner expression alone

Instead of maximizing $f(x)=\left[g(x)\right]^{1/3}$, it is equivalent — and much simpler — to maximize $g(x)$ itself over the same interval, then apply the outer function at the very end.

Step 3: Identify the shape of the inner function and where its extremum lies

If $g(x)$ is a quadratic, write it in the form $ax^2+bx+c$ and note the sign of $a$: a positive $a$ means the parabola opens upward, so its vertex is a minimum, not a maximum.

Step 4: On a closed interval, always compare the vertex with both endpoints …

Common Mistakes

Mistake 1: Assuming the vertex of the quadratic gives the maximum

$g(x)=x^2-x+1$ opens upward, so its vertex at $x=\tfrac12$ is a minimum, not a maximum. A student who reflexively computes the vertex and reports it as "the answer" without checking the shape of the parabola gets the wrong extremum entirely.

Mistake 2: Forgetting that a closed interval's maximum can sit at an endpoint

Even after correctly identifying that the vertex is a minimum, it is easy to forget to actually check both endpoints $x=0$ and $x=1$ — the maximum on $[0,1]$ must come from comparing endpoint values, since there is no other candidate once the vertex is ruled out. …