Problem 40 - Calculus with Concepts in Calculus

Question

Find the intervals on which the given Ir function is increasing and those on which it is decreasing. $$ g(x)=(x-2) /(x-1) $$

Step-by-Step Solution

Answer

The function is increasing on \((-\infty, 1)\) and \((1, \infty)\).

Step 1: Find the Derivative

To determine where the function is increasing or decreasing, we need to find its derivative. We have \( g(x) = \frac{x-2}{x-1} \). We use the quotient rule: if \( u(x) = x-2 \) and \( v(x) = x-1 \), then \( g'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2} \). Calculating each part, \( u'(x) = 1 \), \( v'(x) = 1 \), so \( g'(x) = \frac{(x-1) \cdot 1 - (x-2) \cdot 1}{(x-1)^2} = \frac{x - 1 - x + 2}{(x-1)^2} = \frac{1}{(x-1)^2} \).

Step 2: Determine the Critical Points

The critical points occur where the derivative is zero or undefined. Since \( g'(x) = \frac{1}{(x-1)^2} \), it is never zero because the numerator is a constant (1), not zero anywhere. However, the derivative is undefined at \( x = 1 \) because of division by zero.

Step 3: Analyze Intervals Around Critical Points

We need to consider the intervals around the point \( x=1 \) where the derivative is undefined. We have two intervals: \( (-\infty, 1) \) and \( (1, \infty) \). Since \( g'(x) = \frac{1}{(x-1)^2} \) is positive in both intervals (as the square of any real number is positive), the function is increasing on both intervals.

Step 4: Conclusion About Increasing and Decreasing Intervals

Since the derivative \( g'(x) = \frac{1}{(x-1)^2} \) is positive in both \( (-\infty, 1) \) and \( (1, \infty) \), the function \( g(x) \) is increasing on these intervals and there are no intervals where it is decreasing.

Key Concepts

Increasing and Decreasing Functions

One fundamental concept in calculus is learning how to determine where functions are increasing or decreasing. This is primarily achieved using the derivative. Here's a friendly way to understand it:

The derivative of a function gives us the slope of the tangent line at any given point on the curve. Essentially, it tells us how the function is behaving at that point.

• If the derivative is positive, the function is increasing. Picture the function going uphill, just like the slope of a hill.
• If the derivative is negative, the function is decreasing, which is like going downhill.
• If the derivative is zero, the function might be flat at that point, indicating a potential maximum, minimum, or a "cusp."

By analyzing these slopes, we can map out where portions of the function go up and where they come down, giving us a complete picture of its behavior.

Quotient Rule

The quotient rule is a technique to find the derivative of a function that is the ratio of two differentiable functions. This rule is an essential tool in calculus, and here's how it works in a step-by-step manner:

Suppose we have a function defined as a quotient, like \( g(x) = \frac{u(x)}{v(x)} \). The quotient rule states that the derivative \( g'(x) \) is given by the formula:

\[ g'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2} \]

In this formula:

• \( u(x) \) is the numerator while \( v(x) \) is the denominator function.
• \( u'(x) \) and \( v'(x) \) are their respective derivatives.
• The denominator \( (v(x))^2 \) is always squared, which makes it important to ensure that \( v(x) \) is not zero.

Essentially, this rule helps calculate the change rate of a quotient, keeping in mind the behavior of both the top and bottom functions.

Critical Points

Critical points are vital in understanding a function's overall behavior and graph. They are specific points where the derivative either equals zero or does not exist. Here’s why they are important:

Critical points let us know potential spots for peaks, valleys, and inflection points where the function's direction might change.

• If the derivative is zero at a point, the function might have a local maximum or minimum. Imagine the summit or base of a roller coaster!
• If the derivative does not exist, like in our exercise where \( x=1 \), it indicates a potential discontinuity or vertical tangent.

By identifying critical points, we can segment the function into regions to analyze its behavior, such as increasing or decreasing trends. Understanding these points is essential for sketching the graph and studying the function’s dynamics thoroughly.