Minimizing the Sum of a Number and Four Times Its Square
The problem asks us to find a real number, let's call it \(x\), such that the sum of this number (\(x\)) and four times its square (\(4x^2\)) is the least possible value. We need to minimize the expression:
\(f(x) = 4x^2 + x\)
This is a quadratic expression. We can find the number \(x\) that minimizes this expression using a couple of methods:
Method 1: Using Calculus
- Define the function: We represent the sum as a function of \(x\):
\(f(x) = 4x^2 + x\)
- Find the derivative: To find the minimum or maximum value of a function, we first find its derivative with respect to \(x\). The derivative of \(f(x)\) is denoted as \(f'(x)\).
\(f'(x) = \frac{d}{dx}(4x^2 + x)\)
Using the power rule for differentiation (\(\frac{d}{dx}(x^n) = nx^{n-1}\)), we get:
\(f'(x) = 4(2x) + 1\)
\(f'(x) = 8x + 1\)
- Set the derivative to zero: For the function to have a minimum (or maximum) value, its derivative must be zero at that point.
\(f'(x) = 0\)
\(8x + 1 = 0\)
- Solve for x: Now, we solve this linear equation for \(x\).
\(8x = -1\)
\(x = -\frac{1}{8}\)
- Convert to decimal: To match the options, we convert the fraction to a decimal.
\(x = -0.125\)
- Confirm it's a minimum (Optional): We can use the second derivative test. The second derivative is \(f''(x) = \frac{d}{dx}(8x + 1) = 8\). Since \(f''(x) = 8 > 0\), the function has a local minimum at \(x = -0.125\).
Method 2: Using Properties of Quadratic Functions
The expression \(f(x) = 4x^2 + x\) is a quadratic function in the standard form \(ax^2 + bx + c\), where \(a=4\), \(b=1\), and \(c=0\).
- The graph of a quadratic function is a parabola.
- Since the coefficient of the \(x^2\) term (\(a=4\)) is positive, the parabola opens upwards, meaning it has a minimum value.
- The minimum value occurs at the vertex of the parabola.
- The x-coordinate of the vertex is given by the formula:
\(x = -\frac{b}{2a}\)
- Substitute the values of \(a\) and \(b\):
\(x = -\frac{1}{2 \times 4}\)
\(x = -\frac{1}{8}\)
- Convert the fraction to a decimal:
\(x = -0.125\)
Both methods show that the real number \(x\) for which the sum of the number and four times its square is the least is -0.125.
Final Answer Check
Let's check the value of the expression \(4x^2 + x\) for \(x = -0.125\) (\(-\frac{1}{8}\)):
\(4\left(-\frac{1}{8}\right)^2 + \left(-\frac{1}{8}\right) = 4\left(\frac{1}{64}\right) - \frac{1}{8}\)
\(= \frac{4}{64} - \frac{1}{8} = \frac{1}{16} - \frac{2}{16} = -\frac{1}{16}\)
The minimum value of the sum is \(-1/16\), which occurs when the number \(x\) is \(-1/8\) or -0.125.
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