Problem 42 - Calculus Single Variable

Question

Calculate the first and second derivatives of the given expression, and classify its local extrema. \(x^{2}-2 \log _{8}(x)\)

Step-by-Step Solution

Answer

The expression has a local minimum at \(x = \frac{1}{\sqrt{\ln(8)}}\).

Step 1: Find the First Derivative

The given expression is \(f(x) = x^2 - 2 \log_8(x)\). To find the first derivative, we differentiate each term separately. The derivative of \(x^2\) is \(2x\). For the term \(-2 \log_8(x)\), we convert the logarithm base 8 to natural logarithm, \(-2 \frac{\ln(x)}{\ln(8)}\), and differentiate using the chain rule. The derivative of \(-2 \frac{\ln(x)}{\ln(8)}\) is \(-2 \frac{1}{x \ln(8)}\). Therefore, the first derivative is \(f'(x) = 2x - \frac{2}{x \ln(8)}\).

Step 2: Find Critical Points

To find critical points, set the first derivative equal to zero and solve for \(x\): \(2x - \frac{2}{x \ln(8)} = 0\). Rearranging gives \(2x = \frac{2}{x \ln(8)}\). Multiplying through by \(x\) leads to \(2x^2 = \frac{2}{\ln(8)}\). Solving for \(x\), we get \(x^2 = \frac{1}{\ln(8)}\), hence \(x = \pm \frac{1}{\sqrt{\ln(8)}}\). Since \(x\) must be positive (the logarithm is undefined for non-positive \(x\)), we only consider \(x = \frac{1}{\sqrt{\ln(8)}}\).

Step 3: Find the Second Derivative

To determine the concavity and classify extrema, we need the second derivative. Differentiate \(f'(x) = 2x - \frac{2}{x \ln(8)}\) to get the second derivative. The derivative of \(2x\) is \(2\), and the derivative of \(-\frac{2}{x \ln(8)}\) is \(\frac{2}{x^2 \ln(8)}\), applying the power rule and chain rule. Therefore, the second derivative is \(f''(x) = 2 + \frac{2}{x^3 \ln(8)}\).

Step 4: Classify Local Extrema

Substitute the critical point \(x = \frac{1}{\sqrt{\ln(8)}}\) into the second derivative \(f''(x)\). Since \(f''\left(\frac{1}{\sqrt{\ln(8)}}\right) = 2 + \frac{2}{\left(\frac{1}{\sqrt{\ln(8)}}\right)^3 \ln(8)}\), this value will be positive (since both terms are positive). A positive second derivative implies that the function is concave up at this point, indicating a local minimum. Thus, the critical point \(x = \frac{1}{\sqrt{\ln(8)}}\) is a local minimum.

Key Concepts

Derivatives

Derivatives help us understand how functions change, showing the rate of change at any point. In calculus, the first derivative of a function gives us the slope of the tangent line to the curve at any point.
For instance, with the function we examined, which was given as \(f(x) = x^2 - 2 \log_8(x)\), finding the first derivative involved differentiating each term separately.
The derivative of \(x^2\) is \(2x\), a straightforward application of the power rule.
The logarithmic term involves using the chain rule and understanding logarithms well. This gives:

• First derivative \(f'(x) = 2x - \frac{2}{x \ln(8)}\).

Knowing how to differentiate such terms is crucial for determining the behavior of the function across its domain.

Local extrema

Local extrema are points where a function reaches a local minimum or maximum. These points are essential when analyzing real-world problems because they indicate optimal conditions.
With our function, finding local extrema involves first obtaining the critical points through setting the first derivative to zero. The critical point found was \(x = \frac{1}{\sqrt{\ln(8)}}\).
This critical point helps us identify where these local extreme values can occur in the function.

• Local Minima: Points where the function changes from decreasing to increasing.
• Local Maxima: Points where the function changes from increasing to decreasing.

The determination of the type of extrema (minimum or maximum) requires further analysis using the second derivative, which checks the concavity at that point.

Critical points

Critical points are where the first derivative is zero or undefined. These points are fundamental in finding where a function could reach its highest or lowest values within a local neighborhood.
In the exercise, the critical points were derived by solving \(2x - \frac{2}{x \ln(8)} = 0\). Solving this equation gives us the potential candidates for local minima or maxima.
Here, logarithmic properties and algebraic manipulation play key roles. At these points, the slope of the tangent is zero, suggesting potential turning points for the function.

• They provide important checkpoints in graphing functions and analyzing behavior.
• These are essential for real-world applications in optimization problems.

Understanding critical points helps predict and understand the movement and direction of a function along its graph.

Concavity

Concavity reveals how the curvature of a function appears between critical points on its graph. The second derivative, \(f''(x)\), plays an integral role here.
Concavity tells us whether a function curves upwards or downwards:

• If \(f''(x) > 0\), the graph is concave up, resembling a "U" shape, indicating a local minimum.
• If \(f''(x) < 0\), the graph is concave down, resembling an "\(\cap\)" shape, indicating a local maximum.

For our function, the second derivative was calculated as \(f''(x) = 2 + \frac{2}{x^3 \ln(8)}\). By substituting the critical point, we determined that the second derivative was positive, suggesting a local minimum at \(x = \frac{1}{\sqrt{\ln(8)}}\).
Understanding concavity helps in sketching graphs and provides insights into the dynamic nature of changes exhibited by the function.