Problem 32 - Calculus - AP Edition

Question

Find all values of \(x\) at which the tangent line to the given curve satisfies the stated property. $$y=\frac{x^{2}+1}{x-1} ; \text { horizontal }$$

Step-by-Step Solution

Answer

The values of \(x\) are \(1 + \sqrt{2}\) and \(1 - \sqrt{2}\).

Step 1: Determine the derivative

To find when the tangent line to the curve is horizontal, we need to find the derivative of the function. The derivative of a function gives us the slope of the tangent line. Given the function \( y = \frac{x^2 + 1}{x - 1} \), we use the quotient rule for derivatives. If \( f(x) = \frac{u}{v} \), then \( f'(x) = \frac{u'v - uv'}{v^2} \). Here, \( u = x^2 + 1 \) and \( v = x - 1 \).

Step 2: Apply the Quotient Rule

Differentiate \( u \) and \( v \). We find \( u' = 2x \) and \( v' = 1 \). Applying the quotient rule: \[ y' = \frac{(2x)(x-1) - (x^2+1)(1)}{(x-1)^2} \] Simplify the numerator: \[ y' = \frac{2x^2 - 2x - x^2 - 1}{(x-1)^2} = \frac{x^2 - 2x - 1}{(x-1)^2} \].

Step 3: Set the Derivative to Zero

To find where the tangent is horizontal, set the derivative equal to zero: \[ \frac{x^2 - 2x - 1}{(x-1)^2} = 0 \] This implies: \[ x^2 - 2x - 1 = 0 \].

Step 4: Solve the Quadratic Equation

We solve the quadratic equation \( x^2 - 2x - 1 = 0 \) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1, b = -2, c = -1 \). This gives: \[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-1)}}{2\cdot1} \] \[ x = \frac{2 \pm \sqrt{4 + 4}}{2} \] \[ x = \frac{2 \pm 2\sqrt{2}}{2} \] \[ x = 1 \pm \sqrt{2} \].

Step 5: Check for Domain Restrictions

The function \( y = \frac{x^2 + 1}{x - 1} \) is undefined at \( x = 1 \) because the denominator must not be zero. Thus, the values we found must be within the domain of the function. Therefore, \( x = 1 \pm \sqrt{2} \) are valid solutions.

Key Concepts

Derivative

The concept of a derivative is fundamental in calculus. It helps us understand the rate at which a function is changing at any given point. Essentially, it measures how the output of a function changes as its input changes. For example, if you consider a curve on a graph, the derivative tells you the slope of the tangent line at any point along the curve.
To find the derivative of a function, different rules and techniques are used depending on the structure of the function. It is a crucial step in solving problems related to slopes, velocities, and rates of change. In mathematical notation, the derivative of a function \( f(x) \) is often denoted as \( f'(x) \) or \( \frac{dy}{dx} \).
The goal is to make complex calculations simpler and to facilitate analysis of functions, which can represent diverse real-world situations. Understanding derivatives helps in fields like physics, engineering, and economics, where predicting trends and behavior is necessary.

Quotient Rule

The Quotient Rule is a special technique used for finding the derivative of functions that are divided by one another, specifically when you have a function in the form \( \frac{u}{v} \). The formula for the Quotient Rule is:
\[ f'(x) = \frac{u'v - uv'}{v^2} \]
This formula tells us how to differentiate a function that is expressed as one function over another. In the quotient, \( u \) is the numerator and \( v \) is the denominator. Here, \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \) respectively.
The Quotient Rule is particularly useful when each part of the fraction is a function in itself, for example, \( y = \frac{x^2 + 1}{x - 1} \).
Using this rule correctly requires separately finding the derivatives \( u'\) and \( v'\), then plugging those into the formula. Once you substitute correctly and simplify, you get the desired derivative of the function. This rule is instrumental in calculus as it aids in breaking down more complex problems into manageable parts.

Quadratic Equation

Quadratic equations are polynomial equations of the form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants and \( a eq 0 \). They are quite common in algebra and appear in various real-world scenarios, including physics and economics.
To solve quadratic equations, several methods are available:

• Factoring, when the equation easily breaks into the product of linear factors.
• Completing the square, which involves making the equation a complete square trinomial.
• Using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. This is a universal method that works for any quadratic equation.

The quadratic formula is particularly powerful because it directly provides the solutions of any quadratic equation, even when it cannot be factored easily. In the formula, \( b^2 - 4ac \) is called the discriminant, and it can tell you about the nature of the roots—whether they are real or complex, and if they are distinct or repeated.
Understanding how to solve quadratic equations deepens your grasp of algebra and enhances your problem-solving capabilities in math, providing solutions that apply to a wide array of practical problems.