Problem 35 - University Calculus: Early Transcendentals

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. $$f(x)=\frac{x^{2}-3}{x-2}, x \neq 2$$

Step-by-Step Solution

Answer

Increasing on \((-\infty, 1)\) and \((3, \infty)\); decreasing on \((1, 2)\) and \((2, 3)\). Local max at \( x=1 \), local min at \( x=3 \). No absolute extrema.

Step 1: Find the derivative of the function

To determine where the function \( f(x) = \frac{x^2 - 3}{x - 2} \) is increasing or decreasing, we first need to find its derivative. We'll use the quotient rule: if \( u(x) = x^2 - 3 \) and \( v(x) = x - 2 \), then the derivative \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \). Calculating these gives us \( u'(x) = 2x \), \( v'(x) = 1 \). So, \[ f'(x) = \frac{(2x)(x-2) - (x^2 - 3)(1)}{(x-2)^2} = \frac{2x^2 - 4x - x^2 + 3}{(x-2)^2} = \frac{x^2 - 4x + 3}{(x-2)^2}. \]

Step 2: Find critical points

Set the derivative \( f'(x) = \frac{x^2 - 4x + 3}{(x-2)^2} \) equal to zero to find critical points. The numerator must be zero:\[ x^2 - 4x + 3 = 0. \]Factor this quadratic equation:\[ (x-1)(x-3) = 0, \]yielding critical points \( x = 1 \) and \( x = 3 \). Also note the discontinuity at \( x = 2 \).

Step 3: Determine intervals of increase and decrease

Using the critical points and the discontinuity, break the real number line into intervals: \((-\infty, 1)\), \((1, 2)\), \((2, 3)\), and \((3, \infty)\). Choose test points from each interval (e.g., \(x = 0\), \(x = 1.5\), \(x = 2.5\), \(x = 4\)) and plug them into \( f'(x) \) to determine the sign.- For \( x = 0 \), \( f'(0) = \frac{3}{4} > 0\), so \( f(x) \) is increasing on \((-\infty, 1)\).- For \( x = 1.5 \), \( f'(1.5) = \frac{-0.75}{0.25} < 0\), so \( f(x) \) is decreasing on \((1, 2)\).- For \( x = 2.5 \), \( f'(2.5) = \frac{-0.75}{0.25} < 0\), so \( f(x) \) is decreasing on \((2, 3)\).- For \( x = 4 \), \( f'(4) = \frac{3}{4} > 0\), so \( f(x) \) is increasing on \((3, \infty)\).

Step 4: Identify local and absolute extrema

Check the function values at critical points and behavior near discontinuities:- At \( x = 1\), \( f(x) = \frac{1^2 - 3}{1-2} = 2 \).- At \( x = 3\), \( f(x) = \frac{3^2 - 3}{3-2} = 6 \).Since \( f(x) \) changes from increasing to decreasing at \( x = 1 \), there is a local maximum at \( x = 1 \).Since \( f(x) \) changes from decreasing to increasing at \( x = 3 \), there is a local minimum at \( x = 3 \). For absolute extrema, note behavior as \( x \to \pm \infty \): The function approaches a horizontal asymptote (because the degrees of the numerator and denominator are equal), given by the leading terms: \( y = x \); hence, no absolute maximum or minimum. \( f(x) \) tends towards the linear function's slope, rising or falling without bound.

Key Concepts

Critical Points

In calculus, critical points of a function are where the derivative is zero or undefined. These points help us determine where a function could have a local maximum, minimum, or change direction. They are key in analyzing the function's behavior.

To find critical points for the function \( f(x) = \frac{x^2 - 3}{x - 2} \), we first derived its formula for the derivative \( f'(x) = \frac{x^2 - 4x + 3}{(x-2)^2} \). The critical points occur where this derivative is zero, requiring only the numerator to be zero because the derivative cannot be zero due to the squared term in the denominator.

Solving \( x^2 - 4x + 3 = 0 \), we obtain \((x - 1)(x - 3) = 0\), identifying critical points at \( x = 1 \) and \( x = 3 \).

A discontinuity in the original function at \( x = 2 \) also plays a crucial role. While not a critical point, it indicates where the function is not defined, affecting interval analysis.

Intervals of Increase and Decrease

Intervals where a function is increasing mean that as \( x \) gets larger, \( f(x) \) does too. Conversely, intervals of decrease indicate \( f(x) \) shrinks as \( x \) increases. Finding these intervals helps in understanding the general shape and behavior of a graph.

For \( f(x) = \frac{x^2 - 3}{x - 2} \), using critical points \( x = 1 \), \( x = 3 \), and considering the discontinuity at \( x = 2 \), we divided the real number line into intervals: \((-\infty, 1)\), \((1, 2)\), \((2, 3)\), and \((3, \infty)\).

By selecting test points in each interval and substituting into \( f'(x) \), we determined:

• \((-\infty, 1)\): increasing
  
• \((1, 2)\): decreasing
  
• \((2, 3)\): decreasing
  
• \((3, \infty)\): increasing
  

The function's behavior analysis for increasing and decreasing regions clarifies where the function's graph rises or falls.

Local Extrema

Local extrema are peaks and troughs on a graph where the function temporarily changes from going up to down or vice versa. They're essential for optimizing certain conditions in real-life problems.

For the given function, local extremas are identified at critical points tested against their surrounding intervals. At \( x = 1 \), the function changes from increasing to decreasing, indicating a local maximum. This means around \( x = 1 \), there's a high point on the curve: \( f(1) = 2 \).

Conversely, at \( x = 3 \), the function transitions from decreasing to increasing. This suggests a local minimum, representing a low point at \( x = 3 \) with \( f(3) = 6 \).

Analysis for absolute extrema (overall highest and lowest points across the entire domain) showed none exist due to the function's extended behavior towards infinity, reflecting a horizontal asymptotic pattern.

Quotient Rule

The Quotient Rule is a key calculus tool used to differentiate functions given as one function divided by another. If \( f(x) = \frac{u(x)}{v(x)} \), its derivative is \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \). It's vital for problems involving division of differentiable functions.

In our case, to differentiate \( f(x) = \frac{x^2 - 3}{x - 2} \), we set \( u(x) = x^2 - 3 \) and \( v(x) = x - 2 \). Calculating their derivatives gives \( u'(x) = 2x \) and \( v'(x) = 1 \). Applying the Quotient Rule:

• \( f'(x) = \frac{(2x)(x-2) - (x^2 - 3)(1)}{(x-2)^2} \)
• This simplifies to \( f'(x) = \frac{x^2 - 4x + 3}{(x-2)^2} \)

The Quotient Rule simplifies finding where the slope of the tangent line to the curve is zero or changes, aiding in deeper analysis of the function's behavior.