Problem 44
Question
In Exercises \(35-44,\) find the extreme values of the function and where they occur. $$ y=\frac{x+1}{x^{2}+2 x+2} $$
Step-by-Step Solution
Verified Answer
The function has a local maximum of \(-\frac{1}{2}\) at \(x = -2\).
1Step 1: Find the derivative
To find extreme values, we need the first derivative of the function. Given the function \( y = \frac{x+1}{x^2+2x+2} \), apply the quotient rule: \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \). Here, \( u = x+1 \) and \( v = x^2 + 2x + 2 \). Then, \( u' = 1 \) and \( v' = 2x + 2 \). Substitute these into the formula to find the derivative: \( y' = \frac{(1)(x^2+2x+2) - (x+1)(2x+2)}{(x^2+2x+2)^2} \). Simplify to get \( y' = \frac{x^2 + 2x + 2 - (2x^2 + 4x + 2)}{(x^2+2x+2)^2} = \frac{-x^2 - 2x}{(x^2+2x+2)^2} \).
2Step 2: Set the derivative to zero to find critical points
The critical points occur where the derivative is zero or undefined. Set the numerator of the derivative to zero: \( -x^2 - 2x = 0 \). Factor to find \( -x(x+2) = 0 \), giving \( x = 0 \) and \( x = -2 \). Both are potential critical points because setting the denominator \((x^2+2x+2)^2 = 0\) results in no solutions, as the expression in the square is never zero for real \( x \).
3Step 3: Determine if critical points are extrema
Use the second derivative test or sign changes in the first derivative to classify the critical points. We opt for sign changes in this case. Just before \( x = 0 \, \) test with \( x = -1 \, \), yielding \( y'(-1) = \frac{-1}{3^2} < 0 \). Just after \( x = 0 \,\) test with \( x = 1 \,\), yielding \( y'(1) = \frac{-3}{9^2} < 0 \). No sign change means \( x = 0 \) is not an extremum. Now around \( x = -2 \, \) test with \( x = -3 \) and \( x = -1.5 \). At \( x = -3 \), \( y'(-3) > 0 \) and at \( x = -1.5 \), \( y'(-1.5) < 0 \), indicating a sign change and thus a local maximum at \( x = -2 \).
4Step 4: Calculate the extreme value
Substitute \( x = -2 \) back into the original function to determine the extreme value. \( y = \frac{-2+1}{(-2)^2+2(-2)+2} = \frac{-1}{4 - 4 + 2} = -\frac{1}{2} \).
Key Concepts
Quotient RuleDerivativeCritical PointsSecond Derivative Test
Quotient Rule
When finding the derivative of a quotient of two functions, we need to apply the Quotient Rule. Imagine you have two functions, one on top (numerator) and one at the bottom (denominator). For our function, given by \( y = \frac{x+1}{x^2+2x+2} \), the top function \( u = x+1 \) and the bottom function \( v = x^2 + 2x + 2 \).
The Quotient Rule is expressed as:
To apply this, calculate the derivatives of \( u \) and \( v \). In this problem:
Simplify this expression to get the final derivative form, which is crucial for finding critical points and extrema.
The Quotient Rule is expressed as:
- \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \)
To apply this, calculate the derivatives of \( u \) and \( v \). In this problem:
- \( u' = 1 \) since the derivative of \( x+1 \) is a constant 1.
- \( v' = 2x + 2 \), as the derivative of \( x^2 + 2x + 2 \) is obtained by applying simple power rules.
Simplify this expression to get the final derivative form, which is crucial for finding critical points and extrema.
Derivative
Derivatives are fundamental in calculus and are used to determine how a function changes at any given point. They provide us with the slope of the tangent line to the function's graph at any point.
For the function \( y = \frac{x+1}{x^2+2x+2} \), we use the Quotient Rule to find its derivative. The simplified first derivative is: \ \( y' = \frac{-x^2 - 2x}{(x^2+2x+2)^2} \).
This result tells us how the function's value changes as \( x \) changes.
To find critical points, we set this derivative to zero, determining where the function's slope is zero (i.e., possible extrema points) or where it is undefined (where the denominator is zero). In this case, solving \( -x^2 - 2x = 0 \), we find:
For the function \( y = \frac{x+1}{x^2+2x+2} \), we use the Quotient Rule to find its derivative. The simplified first derivative is: \ \( y' = \frac{-x^2 - 2x}{(x^2+2x+2)^2} \).
This result tells us how the function's value changes as \( x \) changes.
To find critical points, we set this derivative to zero, determining where the function's slope is zero (i.e., possible extrema points) or where it is undefined (where the denominator is zero). In this case, solving \( -x^2 - 2x = 0 \), we find:
- Critical points at \( x = 0 \) and \( x = -2 \).
Critical Points
Critical points occur where the derivative of a function is zero or undefined. These points are essential because they are potential locations for extreme values, which indicate either local maxima, minima, or saddle points.
For the given function, the critical points are found by setting the derivative to zero, \( -x^2 - 2x = 0 \), which gives us:
However, to truly determine if these points are maxima or minima, we must conduct further tests, such as examining sign changes in the derivative or applying the second derivative test.
In our scenario, while \( x = 0 \) does not change signs in the derivative, indicating no extremum, a sign change around \( x = -2 \) reveals a local maximum.
For the given function, the critical points are found by setting the derivative to zero, \( -x^2 - 2x = 0 \), which gives us:
- \( x = 0 \)
- \( x = -2 \)
However, to truly determine if these points are maxima or minima, we must conduct further tests, such as examining sign changes in the derivative or applying the second derivative test.
In our scenario, while \( x = 0 \) does not change signs in the derivative, indicating no extremum, a sign change around \( x = -2 \) reveals a local maximum.
Second Derivative Test
The Second Derivative Test helps classify critical points by examining the concavity of the function. If the second derivative is positive at a critical point, it indicates a local minimum. If negative, it implies a local maximum. If zero, the test is inconclusive.
In our example, however, we used sign changes in the first derivative to determine the nature of the critical points. Let's review why this is helpful.
We check points slightly less and slightly greater than the critical values.
For \( x = 0 \), testing \( x = -1 \) and \( x = 1 \) showed no sign change, confirming no extremum.
However, at \( x = -2 \), testing \( x = -3 \) and \( x = -1.5 \) indicated a sign change from positive to negative, signaling a local maximum.
These tests allow us to draw conclusions about the behavior of the function at critical points, offering insight into its peaks and troughs.
In our example, however, we used sign changes in the first derivative to determine the nature of the critical points. Let's review why this is helpful.
We check points slightly less and slightly greater than the critical values.
For \( x = 0 \), testing \( x = -1 \) and \( x = 1 \) showed no sign change, confirming no extremum.
However, at \( x = -2 \), testing \( x = -3 \) and \( x = -1.5 \) indicated a sign change from positive to negative, signaling a local maximum.
These tests allow us to draw conclusions about the behavior of the function at critical points, offering insight into its peaks and troughs.
Other exercises in this chapter
Problem 44
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