Problem 38 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the interval(s) where the function is increasing and the interval(s) where it is decreasing. $$ f(x)=\ln x^{2} $$

Step-by-Step Solution

Answer

The function \(f(x) = \ln x^{2}\) is increasing on the interval \((0, +\infty)\) and decreasing on the interval \((-\infty, 0)\).

Step 1: Differentiate the function

To find the critical points, we first have to calculate the derivative of function \(f(x) = \ln x^{2}\). By applying the chain rule, we will get: \( f'(x) = \frac{d}{dx} \ln x^{2} = \frac{2x}{x^{2}} = \frac{2}{x} \) Step 2: Find the critical points

Step 2: Determine the critical points

Now, we need to find the critical points - that means the points at which the derivative changes its sign. We will analyze when the derivative (\(f'(x)=\frac{2}{x}\)) is equal to 0 and when it is undefined: \(f'(x)\) is never equal to zero as there is no \(x\) that would make the fraction equal to zero. However, \(f'(x)\) is undefined when the denominator \(x=0\). So the only critical point we have is \(x=0\). Step 3: Find the intervals of increase and decrease

Step 3: Determine when the derivative is positive or negative

Since the only critical point we have is \(x=0\), we need to find the sign of the derivative for values of \(x\) less than 0 and the values of \(x\) more than 0. For \(x < 0\), we have: \(f'(x) = \frac{2}{x}\) Since the denominator is negative, \(f'(x)\) is negative. Thus, the function is decreasing on the interval \((-\infty, 0)\). For \(x > 0\), we have: \(f'(x) = \frac{2}{x}\) Since the denominator is positive, \(f'(x)\) is positive. Thus, the function is increasing on the interval \((0, +\infty)\). Step 4: Write the final answer

Step 4: Intervals of increase and decrease

The function \(f(x) = \ln x^{2}\) is increasing on the interval \((0, +\infty)\) and decreasing on the interval \((-\infty, 0)\).

Key Concepts

Derivative

The derivative is a fundamental concept in calculus, representing the rate at which a function changes at any given point. When we talk about the derivative of a function, we are essentially finding the slope of the tangent line to the graph of the function at any point along its curve.
For the function given in the exercise, \( f(x) = \ln x^2 \), the derivative can be found using calculus rules. By applying the chain rule, we differentiate \( \ln x^2 \) to obtain \( f'(x) = \frac{2x}{x^2} \), which simplifies to \( \frac{2}{x} \). This expression tells us how quickly the function \( f(x) \) is changing at any value of \( x \), excluding where it is not defined.

Increasing and Decreasing Intervals

To determine where a function is increasing or decreasing, we look at the sign of its derivative.
- If \( f'(x) > 0 \), the function is increasing.- If \( f'(x) < 0 \), the function is decreasing.
In our specific case, the derivative \( f'(x) = \frac{2}{x} \) helps us understand the behavior of \( f(x) = \ln x^2 \). For \( x > 0 \), \( f'(x) > 0 \), meaning the function is increasing. For \( x < 0 \), \( f'(x) < 0 \), indicating the function is decreasing. This distinction allows us to determine that the function is decreasing on the interval \((-\infty, 0)\) and increasing on \((0, +\infty)\).

Critical Points

Critical points of a function occur where the derivative is zero or undefined, as these points might indicate a change in the behavior of the function.
In our problem, \( f'(x) = \frac{2}{x} \) is never zero, but it is undefined at \( x = 0 \). This means \( x = 0 \) is a critical point. While \( x = 0 \) does not give us a specific maximum or minimum (as the function does not actually exist at that point due to the logarithm of zero being undefined), it is a boundary where the behavior of the function shifts from decreasing to increasing.

Chain Rule

The chain rule is a pivotal technique in calculus that allows us to differentiate composite functions. It is particularly useful when a function is "inside" another function.
In our function \( f(x) = \ln x^2 \), we are dealing with a composition where the inside function is \( x^2 \) and the outside function is \( \ln \, \). According to the chain rule, if a function \( y = g(u) \) is composed with another function \( u = h(x) \), its derivative is \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).
Applying this to our function, we find \( \frac{d}{dx} \ln x^2 = \frac{1}{x^2} \cdot 2x = \frac{2}{x} \), confirming the derivative calculation is correct and showing the rule's utility in such circumstances.