Problem 36 - Single Variable Calculus

Question

(a) find the intervals on which \(f\) is increasing or decreasing, and (b) find the relative maxima and relative minima of \(\vec{f}\). $$ f(x)=x^{2}-\ln x $$

Step-by-Step Solution

Answer

(a) The function f(x) is decreasing on the intervals \((-∞, -\frac{1}{\sqrt{2}})\) and \((-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\), and increasing on the interval \((\frac{1}{\sqrt{2}}, ∞)\). (b) There is a relative minimum at x = \(\frac{1}{\sqrt{2}}\). There are no relative maxima.

Step 1: Find the derivative of f(x)

First, we need to find the derivative of the given function: \[f(x) = x^2 - \ln x\] By applying the power rule and chain rule, we can find the derivative: \[f'(x) = 2x - \frac{1}{x}\]

Step 2: Find the critical points

Critical points occur when the derivative is zero or undefined. In this case, we need to solve the following equation: \[2x - \frac{1}{x} = 0\] To get rid of the fraction, we can multiply through by x: \[2x^2 - 1 = 0\] Now, we can solve for x by finding the roots of this quadratic equation: \[ x = \pm\frac{1}{\sqrt{2}}\] These are the critical points where the function may have local maxima or local minima.

Step 3: First Derivative Test

We will now apply the first derivative test to determine the intervals of increase and decrease and classify the critical points as local maxima, local minima, or neither. Pick test points in the intervals determined by the critical points (e.g., -1, 0, and 1). Interval 1 (x < -1/√2): - Test point: x = -1 - f'(-1) = 2(-1) - \(-\frac{1}{-1}\) = -3 - f'(x) < 0, this interval is decreasing. Interval 2 (-1/√2 < x < 1/√2): - Test point: x = 0 (undefined, so use x = 0.5) - f'(0.5) = 2(0.5) - \(\frac{1}{0.5}\) = 1 - 2 = -1 - f'(x) < 0, this interval is decreasing. Interval 3 (x > 1/√2): - Test point: x = 1 - f'(1) = 2(1) - \(\frac{1}{1}\) = 2 - 1 = 1 - f'(x) > 0, this interval is increasing. Since f'(x) changes from negative to negative at x = -1/√2, it is neither a local maximum nor a local minimum. However, at x = 1/√2, f'(x) changes from negative to positive, this indicates a local minimum.

Step 4: Final Answer

(a) The function f(x) is decreasing on the intervals \((-∞, -\frac{1}{\sqrt{2}})\) and \((-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\), and increasing on the interval \((\frac{1}{\sqrt{2}}, ∞)\). (b) There is a relative minimum at x = \(\frac{1}{\sqrt{2}}\). There are no relative maxima.

Key Concepts

First Derivative Test

The First Derivative Test is an essential method to analyze the behavior of functions and their critical points. This test allows us to determine intervals where a function is increasing or decreasing and to identify the nature of critical points as local maxima or minima.

To use the First Derivative Test, we follow these steps:

• Identify the critical points of the function by solving where the first derivative is equal to zero or undefined.
• Choose test points from intervals between and beyond these critical points.
• Evaluate the first derivative at these test points to assess the sign of the derivative in each interval.
• If the derivative changes from positive to negative at a critical point, it indicates a local maximum. If it changes from negative to positive, a local minimum is present.
• If the sign does not change, the critical point is classified as neither a maximum nor a minimum.

By applying the First Derivative Test to our function, we can accurately determine the behavior and classify the critical points for deeper insight into the function's graph.

Intervals of Increase and Decrease

Understanding where a function increases or decreases is crucial for sketching its graph and comprehending its general behavior. The intervals of increase and decrease can be found by analyzing the sign of the first derivative, which indicates the slope of the tangent to the curve at any point.

A function is increasing on intervals where its first derivative is positive because the function's slope (or rate of change) is upwards. Conversely, it is decreasing where its derivative is negative, indicating a downward slope.

In our exercise, after finding the derivative to be \(f'(x) = 2x - \frac{1}{x}\), intervals are evaluated:

• In intervals \((-\infty, -\frac{1}{\sqrt{2}})\) and \((-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\), the derivative is negative, showing decreasing behavior.
• In the interval \((\frac{1}{\sqrt{2}}, \infty)\), the derivative is positive, showing increasing behavior.

These intervals help us understand how the function approaches its critical points and the overall direction in which it moves.

Local Maxima and Minima

Local maxima and minima are points where a function reaches a peak or a trough within a neighborhood around that point. These are critical for understanding the function's overall shape and behavior.

To identify local maxima and minima, we use the changes in the sign of the first derivative at the critical points, which help to classify these points.

In our context:

• A **local maximum** occurs when the derivative changes from positive to negative as it passes through a critical point.
• A **local minimum** occurs when the derivative changes from negative to positive.

For the given function, \(f(x) = x^2 - \ln x\), the point at \(x = \frac{1}{\sqrt{2}}\) is a local minimum since the derivative changes from negative to positive at this location. However, there are no local maxima since no point shows a change from positive to negative in the given intervals.

Derivative of a Function

The derivative of a function is a key concept in calculus that provides information about the rate of change of the function at any given point. It tells us how fast something is increasing or decreasing and is represented mathematically as \(f'(x)\), the first derivative.

To find a derivative, we apply the rules of differentiation, such as the power rule, the product rule, the quotient rule, or the chain rule, depending on the function's structure.

For the function given in the exercise, \(f(x) = x^2 - \ln x\), we use:

• The **power rule**: For \(x^n\), the derivative is \(nx^{n-1}\).
• The **chain rule**: Applied for compound functions, i.e., differentiating \(\ln x\), we find its derivative to be \(\frac{1}{x}\).

Thus, the derivative is \(f'(x) = 2x - \frac{1}{x}\). Knowing the derivative allows us to proceed with finding critical points, which guide us in analyzing the local behavior of the function and identifying key features like maxima, minima, or inflection points.