Determine the points of maxima and minima of the function f(x) = 1/8 ln x - bx + x², x 0, where b

Q: Why is the domain $x > 0$?

The function includes a natural logarithm term, $\ln x$. In the real number system, the logarithm is only defined for strictly positive arguments. If $x \le 0$, the function is undefined, so we must restrict our search for maxima and minima to the positive real axis.

Q: Why can't we just use energy conservation?

Energy conservation is a concept from physics, usually applied to systems with forces and potentials. Here, we are dealing with pure calculus and function analysis. We are looking for the stationary points of a mathematical function, which requires the first derivative test, not physical conservation laws.

Q: What happens if $b < 1$?

If $b < 1$, the discriminant of our quadratic equation, $D = 64b^2 - 64$, becomes negative. This means there are no real roots for $f'(x) = 0$. In this case, the derivative is always positive (since the parabola $16x^2 - 8bx + 1$ never touches the x-axis), meaning the function is strictly increasing and has no local maxima or minima.

Question

Determine the points of maxima and minima of the function $f(x) = \frac{1}{8} \ln x - bx + x^2, x > 0$, where $b \ge 0$ is a constant.

Sigma Percentile · Accepted Answer

### Analyzing the Landscape

Imagine you are standing on a landscape defined by the function $f(x) = \frac{1}{8} \ln x - bx + x^2$. This is a story of competing forces.

The logarithmic term $\frac{1}{8} \ln x$ plunges toward negative infinity as $x$ approaches zero, while the quadratic term $x^2$ rockets toward positive infinity as $x$ grows large. The linear term $-bx$ acts as the mediator, shifting the balance of power.

Our mission is to find the peaks and valleys—the local maxima and minima—where the landscape levels off.

### The First Derivative: The Compass of Change

To find where the landscape levels off, we must identify points where the slope is zero. We differentiate $f(x)$ term by term:

$$f'(x) = \frac{d}{dx} \left( \frac{1}{8} \ln x - bx + x^2 ight)$$

The derivative of $\frac{1}{8} \ln x$ is $\frac{1}{8x}$, the derivative of $-bx$ is $-b$, and the derivative of $x^2$ is $2x$. Thus, our compass for the slope is:

$$f'(x) = \frac{1}{8x} - b + 2x$$

### The Quadratic Challenge

Setting $f'(x) = 0$ gives us the equation $\frac{1}{8x} - b + 2x = 0$. Combining these terms over a common denominator of $8x$, we obtain:

$$f'(x) = \frac{1 - 8bx + 16x^2}{8x} = 0$$

For this fraction to be zero, the numerator must be zero. This leads us to the quadratic equation:

$$16x^2 - 8bx + 1 = 0$$

Using the quadratic formula $x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$, we find:

$$x = \frac{8b \pm \sqrt{(-8b)^2 - 4(16)(1)}}{2(16)} = \frac{8b \pm \sqrt{64b^2 - 64}}{32}$$

Simplifying this expression, we arrive at the critical points:

$$x = \frac{b \pm \sqrt{b^2 - 1}}{4}$$

### The Gatekeeper: The Discriminant

The term inside the square root, $b^2 - 1$, acts as the gatekeeper of our existence. If $b^2 - 1 < 0$, there are no real solutions, and the function possesses no extrema.

For the function to exhibit both a maximum and a minimum, we must satisfy the condition **$b \ge 1$**. This ensures the linear term $-bx$ is sufficiently strong to counteract the growth of the other terms.

### The Final Classification: Maxima or Minima?

We have identified two critical points:

$$x_1 = \frac{b - \sqrt{b^2 - 1}}{4} \quad 	ext{and} \quad x_2 = \frac{b + \sqrt{b^2 - 1}}{4}$$

The numerator of our derivative, $16x^2 - 8bx + 1$, represents an upward-opening parabola. It is positive outside the roots and negative between them.

Therefore, at **$x_1$ (the smaller root)**, the derivative changes from positive to negative, indicating a **local maximum**. At **$x_2$ (the larger root)**, the derivative changes from negative to positive, indicating a **local minimum**.

Animated Solution for Mathematics - Differentiation: Determine the points of maxima and minima of the function $f (x) = \frac{1}{8} ln x - b x + x^{2}, x > 0$ , where $b \geq 0$ is a constant.

Visualized Solution

Understanding the Function $f (x)$

Condition for Maxima and Minima

Finding the First Derivative $f^{'} (x)$

Simplifying $f^{'} (x)$

Finding Critical Points

Solving for $x$

Simplifying the Roots

Condition for Extrema to Exist

Classifying the Critical Points

Final Answer

The Sigma Insight: Maxima and Minima

Analyzing the Landscape

The First Derivative

The Quadratic Challenge

The Gatekeeper

The Final Classification

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