What is the minimum value of [x(x – 1) + 1] 1/3 , where 0 ≤ x ≤ 1?
Finding the Minimum Value of the Function on a Closed Interval
The problem asks for the minimum value of the function $f(x) = [x(x – 1) + 1]^{1/3}$ on the closed interval $0 \le x \le 1$. To find the minimum value of this function, we first look at the expression inside the cube root.
Let $g(x) = x(x – 1) + 1$. Simplifying this expression, we get:
\(g(x) = x^2 - x + 1\)
So, the original function can be written as \(f(x) = [g(x)]^{1/3}\). The cube root function, \(h(y) = y^{1/3}\), is an increasing function. This means that the minimum value of \(f(x)\) will occur at the same point where \(g(x)\) has its minimum value.
Minimizing the Quadratic Function \(g(x)\)
We need to find the minimum value of the quadratic function \(g(x) = x^2 - x + 1\) on the interval \(0 \le x \le 1\).
A quadratic function in the form \(ax^2 + bx + c\) has a parabola shape. Since the coefficient of \(x^2\) is positive (a=1), the parabola opens upwards, and its minimum value occurs at the vertex.
The x-coordinate of the vertex of a parabola \(ax^2 + bx + c\) is given by the formula \(x = -b / (2a)\).
For \(g(x) = x^2 - x + 1\), we have \(a=1\) and \(b=-1\).
The x-coordinate of the vertex is:
\(x = -(-1) / (2 \times 1) = 1 / 2\)
Evaluating at the Vertex and Endpoints
The interval given is \(0 \le x \le 1\). The vertex we found, \(x = 1/2\), lies within this interval.
To find the minimum value of \(g(x)\) on the closed interval, we need to evaluate \(g(x)\) at the vertex and at the endpoints of the interval:
- Vertex: \(x = 1/2\)
- Left endpoint: \(x = 0\)
- Right endpoint: \(x = 1\)
Value of \(g(x)\) at the Vertex (\(x = 1/2\))
\(g(1/2) = (1/2)^2 - (1/2) + 1 = 1/4 - 1/2 + 1 = 1/4 - 2/4 + 4/4 = (1 - 2 + 4) / 4 = 3/4\)
Value of \(g(x)\) at the Left Endpoint (\(x = 0\))
\(g(0) = 0^2 - 0 + 1 = 0 - 0 + 1 = 1\)
Value of \(g(x)\) at the Right Endpoint (\(x = 1\))
\(g(1) = 1^2 - 1 + 1 = 1 - 1 + 1 = 1\)
Comparing Values to Find the Minimum of \(g(x)\)
We compare the values of \(g(x)\) at the vertex and endpoints:
- \(g(1/2) = 3/4\)
- \(g(0) = 1\)
- \(g(1) = 1\)
The minimum value of \(g(x)\) on the interval \(0 \le x \le 1\) is the smallest of these values, which is \(3/4\).
Finding the Minimum Value of \(f(x)\)
Since \(f(x) = [g(x)]^{1/3}\) and \(g(x)\) has a minimum value of \(3/4\) on the interval, the minimum value of \(f(x)\) is:
\(f_{\text{min}} = [g_{\text{min}}]^{1/3} = [3/4]^{1/3}\)
\(f_{\text{min}} = {\left( {\frac{3}{4}} \right)^{\frac{1}{3}}}\)
This corresponds to one of the given options.
Revision Table: Key Steps
| Step | Description | Result |
|---|---|---|
| 1 | Simplify the function inside the cube root. | \(g(x) = x^2 - x + 1\) |
| 2 | Identify function type and method to find minimum. | Quadratic, find vertex and check endpoints. |
| 3 | Calculate vertex x-coordinate of \(g(x)\). | \(x = 1/2\) |
| 4 | Verify vertex is within the interval [0, 1]. | Yes, \(1/2\) is in [0, 1]. |
| 5 | Evaluate \(g(x)\) at vertex and endpoints. | \(g(1/2) = 3/4\), \(g(0) = 1\), \(g(1) = 1\) |
| 6 | Find the minimum value of \(g(x)\) on the interval. | \(g_{\text{min}} = 3/4\) |
| 7 | Calculate the minimum value of \(f(x)\). | \(f_{\text{min}} = (3/4)^{1/3}\) |
Additional Information: Finding Extrema on a Closed Interval
To find the absolute maximum and minimum values of a continuous function \(f(x)\) on a closed interval \([a, b]\), you follow these steps:
- Find the critical points of \(f(x)\) within the interval \((a, b)\). Critical points are where \(f'(x) = 0\) or \(f'(x)\) is undefined.
- Evaluate \(f(x)\) at each critical point found in step 1.
- Evaluate \(f(x)\) at the endpoints of the interval, \(a\) and \(b\).
- Compare all the values obtained in steps 2 and 3. The largest value is the absolute maximum on \([a, b]\), and the smallest value is the absolute minimum on \([a, b]\).
In this specific problem, because \(f(x)\) is a composition of an increasing function (\(y^{1/3}\)) and a differentiable function (\(g(x)\)), we could minimize \(g(x)\) first. \(g(x)\) is a simple quadratic, and its minimum on an interval is either at its vertex (if the vertex is in the interval) or at one of the endpoints.
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