Understanding the Problem

We are asked to find the non-negative values of a parameter '\(b\)' such that the function \(f(x) = \frac{16x^3}{3} - 4bx^2 + x\) does not have any local maximum or local minimum points in the domain where \(x > 0\). A key condition is that '\(b\)' must be non-negative, meaning \(b \ge 0\).

Function and Its Derivative

The given function is:

\(f(x) = \frac{16x^3}{3} - 4bx^2 + x\)

To find the points where the function might have a maximum or minimum, we first need to calculate its first derivative with respect to \(x\):

\(f'(x) = \frac{d}{dx} \left( \frac{16x^3}{3} - 4bx^2 + x \right)\)

\(f'(x) = \frac{16 \cdot 3x^2}{3} - 4b \cdot 2x + 1\)

\(f'(x) = 16x^2 - 8bx + 1\)

Analyzing Critical Points

Local maxima and minima occur at critical points, where the first derivative \(f'(x)\) is equal to zero or is undefined. Since \(f'(x)\) is a polynomial, it is defined for all \(x\). Thus, we only need to consider points where \(f'(x) = 0\).

Setting the derivative to zero gives a quadratic equation:

\(16x^2 - 8bx + 1 = 0\)

The function \(f(x)\) will have neither a maximum nor a minimum in the range \(x > 0\) if its derivative \(f'(x)\) does not change sign for \(x > 0\). A common interpretation to ensure this is that \(f'(x)\) must never be equal to zero within the specified range (\(x > 0\)).

Evaluating the Discriminant

To understand the roots of the quadratic equation \(16x^2 - 8bx + 1 = 0\), we calculate its discriminant (\(\Delta\)):

\(\Delta = (\text{coefficient of } x)^2 - 4 \times (\text{coefficient of } x^2) \times (\text{constant term})\)

\(\Delta = (-8b)^2 - 4(16)(1)\)

\(\Delta = 64b^2 - 64\)

\(\Delta = 64(b^2 - 1)\)

Determining Conditions for No Extrema

We analyze the nature of the roots based on the discriminant and the requirement that \(f'(x) \ne 0\) for \(x > 0\). Remember that \(b \ge 0\).

• Case 1: No real roots (\(\Delta < 0\))

  If \(\Delta < 0\), the quadratic equation \(16x^2 - 8bx + 1 = 0\) has no real roots. Since the leading coefficient (16) is positive, the parabola representing \(f'(x)\) opens upwards and is always above the x-axis. This means \(f'(x) > 0\) for all \(x\). If \(f'(x)\) is always positive, the function \(f(x)\) is strictly increasing and has no local maxima or minima.

  The condition \(\Delta < 0\) translates to:

  \(64(b^2 - 1) < 0\)

  \(b^2 - 1 < 0\)

  \(b^2 < 1\)

  This inequality holds for \(-1 < b < 1\). Since we are given \(b \ge 0\), this case applies when \(0 \le b < 1\).

• Case 2: One real root (\(\Delta = 0\))

  If \(\Delta = 0\), the quadratic equation has exactly one real root (a repeated root). This occurs when:

  \(64(b^2 - 1) = 0\)

  \(b^2 = 1\)

  Since \(b \ge 0\), we get \(b = 1\).

  In this case (\(b = 1\)), the derivative becomes \(f'(x) = 16x^2 - 8x + 1 = (4x - 1)^2\).

  The equation \(f'(x) = 0\) yields \((4x - 1)^2 = 0\), which has a single root \(x = 1/4\). Since \(1/4 > 0\), this root lies within our specified range \(x > 0\). At this point \(x = 1/4\), the derivative \(f'(x)\) is zero. If the requirement is strictly that \(f'(x)\) must *never* be zero for \(x > 0\) to ensure no extrema, then \(b=1\) must be excluded.

• Case 3: Two distinct real roots (\(\Delta > 0\))

  If \(\Delta > 0\), the quadratic equation has two distinct real roots. This occurs when:

  \(64(b^2 - 1) > 0\)

  \(b^2 - 1 > 0\)

  \(b^2 > 1\)

  Since \(b \ge 0\), this condition implies \(b > 1\).

  When \(b > 1\), the roots of \(f'(x) = 0\) are \(x_{1,2} = \frac{8b \pm \sqrt{64b^2 - 64}}{32} = \frac{b \pm \sqrt{b^2 - 1}}{4}\). Both roots are positive. Because \(f'(x)\) is a quadratic opening upwards, it will be positive outside these roots and negative between them. This change in sign of \(f'(x)\) indicates that the function \(f(x)\) has a local maximum at the smaller root (\(x_1\)) and a local minimum at the larger root (\(x_2\)). Both occur in the range \(x > 0\), so this case does not satisfy the condition.

Final Determination for \(b\)

Based on the analysis, the function \(f(x)\) has neither a maximum nor a minimum in the range \(x > 0\) only when the derivative \(f'(x)\) is strictly positive (never zero) for all \(x > 0\). This occurs only when the discriminant \(\Delta\) is negative.

The condition \(\Delta < 0\) leads to \(0 \le b < 1\).

Therefore, the non-negative values of \(b\) for which the function has neither maximum nor minimum in the range \(x > 0\) are \(0 \le b < 1\).