Problem 35 - Thomas Calculus in SI Units

Question

a. Find the open intervals on which the function is increasing and decreasing. b. Identify the function's local and absolute extreme values, if any, saying where they occur. $$f(x)=\frac{x^{2}-3}{x-2}, \quad x \neq 2$$

Step-by-Step Solution

Answer

The function increases on \((-\infty, 1)\), \((2, 3)\), and \((3, \infty)\); decreases on \((1, 2)\); and has a local max at \( x = 1 \) and a local min at \( x = 3 \).

Step 1: Find the Derivative

To determine where the function is increasing or decreasing, we first need to find the derivative of the function.Given the function:\[ f(x) = \frac{x^2 - 3}{x - 2} \]Using the quotient rule for derivatives, \( \frac{u}{v} \), where \( u = x^2 - 3 \) and \( v = x - 2 \), the derivative \( f'(x) \) is given by:\[ f'(x) = \frac{(x - 2)(2x) - (x^2 - 3)(1)}{(x - 2)^2} \]This simplifies to:\[ f'(x) = \frac{2x^2 - 4x - x^2 + 3}{(x - 2)^2} = \frac{x^2 - 4x + 3}{(x - 2)^2} \]

Step 2: Find Critical Points

To find critical points, set \( f'(x) = 0 \) or where the derivative does not exist (undefined).Solve \( x^2 - 4x + 3 = 0 \) by factoring to find values that make the numerator zero:\[ (x-3)(x-1) = 0 \]Thus, \( x = 3 \) and \( x = 1 \) are critical points. The derivative is also undefined when \( x = 2 \) because this makes the denominator zero.

Step 3: Determine Intervals of Increase and Decrease

To determine if the function is increasing or decreasing in the intervals defined by the critical points and where the derivative is undetermined, evaluate \( f'(x) \) at test points:- Interval \( (-\infty, 1) \): Choose \( x = 0 \) and compute \( f'(0) = \frac{3}{4} \) (positive).- Interval \( (1, 2) \): Choose \( x = 1.5 \) and compute \( f'(1.5) = \frac{-0.25}{0.25} \) (negative).- Interval \( (2, 3) \): Choose \( x = 2.5 \) and compute \( f'(2.5) = \frac{1.75}{0.25} \) (positive).- Interval \( (3, \infty) \): Choose \( x = 4 \) and compute \( f'(4) = \frac{3}{1} \) (positive).This shows the function is increasing on \((-\infty, 1)\), \((2, 3)\), and \((3, \infty)\), and decreasing on \((1, 2)\).

Step 4: Determine Local and Absolute Extrema

Examine \( f(x) \) at critical points (\( x = 1 \) and \( x = 3 \)) and where \( x = 2 \):- At \( x = 1 \), the function transitions from increasing to decreasing, indicating a local maximum.- At \( x = 2 \), the function is undefined, no extrema.- At \( x = 3 \), the function transitions from decreasing to increasing, indicating a local minimum.Evaluate the function at these critical points:\[ f(1) = \frac{1^2 - 3}{1 - 2} = 2 \] (local maximum)\[ f(3) = \frac{3^2 - 3}{3 - 2} = 6 \] (local minimum)There are no absolute extrema as \( x \to \infty \) and \( x \to -\infty \).

Step 5: Conclude on Function Behavior

The function \( f(x) = \frac{x^2 - 3}{x - 2} \) increases on intervals \((-\infty, 1)\), \((2, 3)\), and \((3, \infty)\), and decreases on \((1, 2)\). It has a local maximum at \( x = 1 \) with a value of 2, and a local minimum at \( x = 3 \) with a value of 6.

Key Concepts

Derivatives

To understand how a function behaves, we often turn to derivatives. Imagine derivatives as tools that show us the slope or steepness of a curve at any point. For the given function \( f(x) = \frac{x^2 - 3}{x - 2} \), the derivative, found using the quotient rule, helps us identify where the function increases or decreases. The quotient rule is especially useful here because our function is a fraction, with a polynomial in both the numerator and the denominator.
The quotient rule states that for a function \( u/v \), the derivative is \( f'(x) = \frac{v(u') - u(v')}{v^2} \). This rule allows you to calculate how the change in the numerator affects the whole function in relation to its denominator, giving us \( f'(x) = \frac{x^2 - 4x + 3}{(x - 2)^2} \). Understanding derivatives is crucial for dissecting how a curve behaves, identifying possible peaks and troughs along its path.

Critical Points

Critical points are like the key highlights on a function's journey, marking where special action happens. These are the \( x \) values where the derivative \( f'(x) = 0 \) or is undefined, indicating a possible change in the function's direction. In our example, solving \( x^2 - 4x + 3 = 0 \) leads us to \( x = 1 \) and \( x = 3 \). Additionally, the derivative is undefined at \( x = 2 \), since it makes the denominator zero and thus the whole expression undefined.
These critical points are important because they mark potential local maxima and minima, or places where the function shifts its slope from positive (increasing) to negative (decreasing), or vice versa. Understanding where these points lie helps us paint a clearer picture of the function's overall behavior.

Extrema

In calculus, extrema are the "peaks and valleys" of a function, represented by its highest and lowest values. These include both local extrema, which are points of maximum or minimum values within a certain interval, and absolute extrema, which are the highest or lowest points in the entire function.
For the function \( f(x) = \frac{x^2 - 3}{x - 2} \), we find a local maximum at \( x = 1 \) with the value 2, and a local minimum at \( x = 3 \) with the value 6. These occur where the function's behavior changes from increasing to decreasing (or vice versa). Extrema are essential for understanding the shape of a function, because they tell us how it caps off or bottoms out at different points.

Increasing and Decreasing Intervals

When analyzing a function, determining the intervals of increase and decrease helps us understand its general trend. This involves checking the sign of the derivative across different sections of the graph. For \( f(x) = \frac{x^2 - 3}{x - 2} \), if \( f'(x) > 0 \), the function is increasing in that interval. Conversely, if \( f'(x) < 0 \), the function is decreasing.
For instance, analyzing around the critical points, we find that the function increases on intervals \((-\infty, 1)\), \((2, 3)\), and \((3, \infty)\), but decreases on \((1, 2)\). This evaluation of increasing and decreasing intervals provides insight into the function's momentum and direction, much like understanding when a car speeds up or slows down on a hilly road.