Problem 133 - Calculus Volume 1

Question

For the following exercises, find the local and absolute minima and maxima for the functions over \((-\infty, \infty)\) $$y=\frac{x^{2}+x+6}{x-1}$$

Step-by-Step Solution

Answer

Local extrema at \( x = \pm\sqrt{7} \); no absolute extrema.

Step 1: Find the Critical Points

First, we need to find the derivative of the function, which will help us find the critical points. The function is \( y = \frac{x^2 + x + 6}{x - 1} \). We use the quotient rule to differentiate: \( f(x) = x^2 + x + 6 \) and \( g(x) = x - 1 \). The quotient rule is \( \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2} \).So, \( f'(x) = 2x + 1 \) and \( g'(x) = 1 \). The derivative \( y' = \frac{(2x+1)(x-1) - (x^2+x+6)(1)}{(x-1)^2} \).Simplify to find the critical points:\( y' = \frac{2x^2 - 2x + x - 1 - x^2 - x - 6}{(x-1)^2} = \frac{x^2 - 7}{(x-1)^2} \).Setting \( y' = 0 \), we solve \( x^2 - 7 = 0 \) to find the critical points \( x = \pm \sqrt{7} \).

Step 2: Determine the Nature of Critical Points

To determine whether these critical points are local minima or maxima, we use the second derivative test. First, we find the second derivative of the function. The second derivative is derived from the first derivative \( y' = \frac{x^2 - 7}{(x-1)^2} \).Find \( y'' \) and evaluate at the critical points \( x = \sqrt{7} \) and \( x = -\sqrt{7} \):Calculating, \( y'' = \frac{d}{dx} \left( \frac{x^2 - 7}{(x-1)^2} \right) \) using quotient or chain rule, checking the sign will provide if it's a local minima (positive) or maxima (negative).

Step 3: Evaluate the Function at Critical Points

Evaluate the function at \( x = \sqrt{7} \) and \( x = -\sqrt{7} \) using the original function:\( y(\sqrt{7}) = \frac{(\sqrt{7})^2 + \sqrt{7} + 6}{\sqrt{7} - 1} \).Simplifying gives you the value of the function at this critical point. Repeat for \( x = -\sqrt{7} \).These evaluations will confirm the relative values (whether local minima or maxima).

Step 4: Consider the End Behavior

Since the function is defined over \((-\infty, \infty)\), consider the limits as \( x \to \infty \) and \( x \to -\infty \). By inspecting the leading terms, as \( x \to \infty \), the function behaves approximately as \( y \equiv x^2 / x = x \to \infty \), and \( x \to -\infty \), \( y \to -x \to -\infty \).Check also what happens as \( x \to 1\), the denominator makes \( y \to \infty \) indicating no absolute extrema at these asymptotes.

Step 5: Conclude on Local and Absolute Extrema

Given the second derivative test and the function evaluations, we determine:- The critical points \( x = \sqrt{7} \) or \( x = -\sqrt{7} \) give the local minima and maxima if confirmed by the second derivative signs.- Due to the function's growth outward both positive and negative, there are no absolute maxima or minima across all real numbers. The end behavior dominates as \( |x| \to \infty \).

Key Concepts

Derivative Calculation

Calculating derivatives is a fundamental skill in calculus, as it allows us to find rates of change and slopes of curves. In the exercise provided, we have the function \[ y = \frac{x^2 + x + 6}{x - 1} \]To find its derivative, we use the Quotient Rule. This method is essential when dealing with functions that are written as one function divided by another.Let's break it down:

• The Quotient Rule states: \( \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2} \), where \( f(x) \) is the numerator and \( g(x) \) is the denominator.
• For our function, \( f(x) = x^2 + x + 6 \) and \( g(x) = x - 1 \).
• We first find \( f'(x) \) and \( g'(x) \): \( f'(x) = 2x + 1 \) and \( g'(x) = 1 \).

This leads to the derivative \\[ y' = \frac{(2x+1)(x-1) - (x^2+x+6)(1)}{(x-1)^2} \]By simplifying, we find:\[ y' = \frac{x^2 - 7}{(x-1)^2} \]Understanding how to apply and simplify using the Quotient Rule is a key part of learning calculus.

Critical Points

Critical points are where the derivative of a function equals zero or is undefined. These points help us identify peaks, troughs, or flat sections in the graph of a function, and they're critical (pun intended) in determining local extrema.For our function, we set the derivative \( y' = \frac{x^2 - 7}{(x-1)^2} \) to zero to find any such points:

• We solve \( x^2 - 7 = 0 \), which gives us \( x = \pm \sqrt{7} \).
• These solutions represent the critical points of the function.

To be thorough, it's worth noting that the critical points are locations where the slope is either zero (indicating a potential local max or min) or undefined.In this exercise, the function is undefined for \( x = 1 \), but \( x = \pm\sqrt{7} \) are our valid critical points. Finding these points is the first step to analyzing the behavior of the function.

Local and Absolute Extrema

Once we have the critical points, it's time to find out if they are local max/min points. Extrema refer to these maximum or minimum points of a function, and they can be local or absolute.For this function:

• Local extrema occur at critical points within the interval, where the function's slope changes sign.
• In our case, we'll evaluate the function at \( x = \sqrt{7} \) and \( x = -\sqrt{7} \) to ascertain their nature.

An absolute extremum is the highest or lowest value of the function over its entire domain.

• Given our function is defined on \((-\infty, \infty)\) and its behavior is asymptotic as it approaches infinity, it lacks absolute extrema over the reals because it doesn’t reach a highest or lowest point as \( x \to \infty \) or \( x \to -\infty \).
• The behavior around \( x = 1 \) potentially also creates discontinuities, so no absolute extrema there either.

Understanding the difference between local and absolute extrema gives insight into the overall tendencies and extremities of a function.

Second Derivative Test

The Second Derivative Test provides a way to determine whether a critical point is a local maximum, local minimum, or a saddle point by examining the second derivative.Here’s how it works in this problem:

• We first compute the second derivative \( y'' \) based on \( y' = \frac{x^2 - 7}{(x-1)^2} \).
• If \( y''(x) > 0 \) at a critical point, the function has a local minimum at that point.
• If \( y''(x) < 0 \), it indicates a local maximum.
• If \( y''(x) = 0 \), the test is inconclusive, and other methods must be used to ascertain the behavior.

In this exercise, analyzing \( y'' \) at \( x = \pm \sqrt{7} \) provides whether these points are indeed local minima or maxima. The second derivative thus acts as a useful marker for understanding a function's curvature at critical points, underpinning the study of concavity and extremality.