Chapter 3: Problem 44
Solve the equation \(D_{x} y=0\). \(y=\frac{x^{2}+2 x+5}{x+1}\)
Short Answer
Expert verified
The derivative is zero at \( x = -3 \) and \( x = 1 \).
Step by step solution
01
Understand the Problem
We need to find the first derivative of the function \( y = \frac{x^{2} + 2x + 5}{x + 1} \) and then determine the points where the derivative equals zero.
02
Identify the Differentiation Method
The function is a rational function, so we will use the quotient rule for differentiation. The quotient rule states that if \( y = \frac{u}{v} \), then \( y' = \frac{u'v - uv'}{v^2} \), where \( u = x^2 + 2x + 5 \) and \( v = x + 1 \).
03
Differentiate the Numerator and Denominator
Differentiate \( u = x^2 + 2x + 5 \) to get \( u' = 2x + 2 \). Differentiate \( v = x + 1 \) to get \( v' = 1 \).
04
Apply the Quotient Rule
Using the quotient rule, the derivative \( y' = \frac{(2x + 2)(x+1) - (x^2 + 2x + 5)(1)}{(x+1)^2} \). Simplify the expression: \( y' = \frac{(2x^2 + 2x + 2x + 2) - (x^2 + 2x + 5)}{(x+1)^2} \).
05
Simplify the Expression
Simplifying the numerator: \( 2x^2 + 4x + 2 - x^2 - 2x - 5 = x^2 + 2x - 3 \). Therefore, \( y' = \frac{x^2 + 2x - 3}{(x+1)^2} \).
06
Set the Derivative Equal to Zero
We need to solve \( \frac{x^2 + 2x - 3}{(x+1)^2} = 0 \). For a fraction to be zero, its numerator must be zero, so solve \( x^2 + 2x - 3 = 0 \).
07
Solve the Quadratic Equation
Factor the quadratic equation: \( x^2 + 2x - 3 = (x + 3)(x - 1) = 0 \). Thus, the solutions are \( x = -3 \) and \( x = 1 \).
08
Verify the Solutions
Substitute \( x = -3 \) and \( x = 1 \) back into the original function to ensure these points make the derivative zero. Both points are valid.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quotient Rule
The quotient rule is a technique used in calculus for differentiating functions that are the ratio of two differentiable functions. When facing a problem where you need to differentiate a rational function, the quotient rule is your tool of choice.
Here's the rule in a nutshell: If you have a function represented as \( y = \frac{u}{v} \), where \( u \) and \( v \) are differentiable functions of \( x \), the derivative of \( y \), denoted as \( y' \), is given by:
Here's the rule in a nutshell: If you have a function represented as \( y = \frac{u}{v} \), where \( u \) and \( v \) are differentiable functions of \( x \), the derivative of \( y \), denoted as \( y' \), is given by:
- \( y' = \frac{u'v - uv'}{v^2} \)
- Differentiate the numerator \( u \), denoted \( u' \).
- Differentiate the denominator \( v \), denoted \( v' \).
- Multiply \( u' \) by \( v \) and subtract the product of \( u \) and \( v' \).
- Divide the result by the square of \( v \) (i.e., \( v^2 \)).
Rational Function
A rational function is a type of function that can be expressed as the quotient of two polynomials. In other words, it's a fraction with polynomials in both the numerator and the denominator. This kind of function appears often in calculus exercises.
Here's a concise picture of what rational functions involve:
Here's a concise picture of what rational functions involve:
- They are expressed in the form \( f(x) = \frac{p(x)}{q(x)} \), where \( p(x) \) and \( q(x) \) are polynomials and \( q(x) eq 0 \).
- They can have a mix of constant terms and terms with varying powers of \( x \).
- Understanding the behavior of rational functions is crucial, especially when working with calculus as they often require differentiation using the quotient rule.
Derivative
The derivative is a fundamental concept in calculus that represents the rate at which a function is changing at any given point. It's essentially the slope of the function at each specific point. When determining the derivative of a function like the one given, you're evaluating how the function's output changes as its input changes.
Key aspects of derivatives include:
Key aspects of derivatives include:
- The derivative of a function \( f(x) \) is often denoted as \( f'(x) \) or \( \frac{dy}{dx} \).
- It provides crucial information about turning points, slopes, and the function's behavior and trends.
- Once the derivative is determined, you can set it to zero to find critical points, which often represent maxima, minima, or inflection points in the context of functions.
Quadratic Equation
A quadratic equation is a specific type of polynomial equation of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). These equations are fundamental in the study of mathematics, and they frequently occur when solving derivative problems involving second-degree polynomials as seen in this exercise.
Solving quadratic equations can be approached in several ways:
Solving quadratic equations can be approached in several ways:
- **Factoring:** Express the equation in a product of its factors, like in our exercise: \( (x + 3)(x - 1) = 0 \).
- **Quadratic Formula:** Use the formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) when the equation can't be easily factored.
- **Completing the Square:** Rewriting the equation in a squared binomial form to easily solve for the variable \( x \).