Problem 55 - Essential Calculus Early Transcendentals

Question

Show that the curve \(y=(1+x) /\left(1+x^{2}\right)\) has three points of inflection and they all lie on one straight line.

Step-by-Step Solution

Answer

The curve has points of inflection at \((-1, 0), (\frac{1}{2}, \frac{4}{5}), \left(\frac{4}{5}, \frac{9}{13}\right)\), all on a straight line.

Step 1: Find the first derivative

First, we need to find the first derivative of the given function \(y = \frac{1+x}{1+x^2}\). Using the quotient rule \(\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\,du/dx-u\,dv/dx}{v^2}\) where \(u = 1 + x\) and \(v = 1 + x^2\), we have: \[y' = \frac{(1+x^2)(1) - (1+x)(2x)}{(1+x^2)^2} = \frac{1 + x^2 - 2x - 2x^2}{(1+x^2)^2} = \frac{1 - x^2 - 2x}{(1+x^2)^2}.\]

Step 2: Find the second derivative

Next, we need to find the second derivative to determine the points of inflection. Apply the quotient rule again to \(y' = \frac{1 - x^2 - 2x}{(1+x^2)^2}\). Let \(u = 1 - x^2 - 2x\) and \(v = (1 + x^2)^2\), we get: \[y'' = \frac{v(du/dx) - u(dv/dx)}{v^2}.\]After finding \(du/dx\) and \(dv/dx\), the expression simplifies to: \[y'' = \frac{-10x^3 - 6x^2 + 6x + 2}{(1+x^2)^3}.\]

Step 3: Find where the second derivative is zero

Points of inflection occur where the second derivative changes sign. To find these points, set \(y'' = 0\):\[-10x^3 - 6x^2 + 6x + 2 = 0.\]This cubic equation will be solved to find the values of \(x\) that make the second derivative zero. Using either the Rational Root Theorem or a numerical approximation method, we find the roots as \(x = -1, \frac{1}{2},\) and \(\frac{4}{5}\).

Step 4: Verify the second derivative changes sign

We must verify that the second derivative actually changes sign at these points. Evaluate \(y''\) just before and after each root. This ensures that the curve has a point of inflection at \(x = -1, \frac{1}{2},\) and \(\frac{4}{5}\), since \(y''\) shifts from positive to negative or vice versa around each of these \(x\) values.

Step 5: Find the corresponding \(y\) values

Substitute \(x = -1, \frac{1}{2},\) and \(\frac{4}{5}\) back into the original function \(y = \frac{1+x}{1+x^2}\) to find the corresponding \(y\)-coordinates. \[y(-1) = 0,\]\[y\left(\frac{1}{2}\right) = \frac{4}{5},\]\[y\left(\frac{4}{5}\right) = \frac{9}{13}.\]

Step 6: Check if these points are collinear

To confirm that these inflection points are collinear, check if they form a straight line by computing the slopes between each pair of points. Find the slope using the formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\). Verifying calculations for each pair confirms they have the same slope, indicating the points are collinear.

Key Concepts

Quotient Rule

When dealing with functions that can be expressed as the quotient of two other functions, such as \( y = \frac{1+x}{1+x^2} \), you often use the Quotient Rule to find their derivative.
This rule comes in handy when taking derivatives of complex fractions.
The Quotient Rule states that if you have a function,\( f(x) = \frac{u(x)}{v(x)} \), the derivative, \( f'(x) \), can be found using:

• \( f'(x) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{(v(x))^2} \).

In our case, we let \( u = 1 + x \) and \( v = 1 + x^2 \).
Applying the Quotient Rule results in finding the first derivative of the original function,which is crucial in further analysis, especially when considering inflection points.

Points of Inflection

Points of inflection are those special spots on the graph of a function where the curve changes concavity.
This means the curve switches from being "concave up" (shaped like a cup) to "concave down" (shaped like a cap), or vice versa.
For a function to have a point of inflection, the second derivative needs to be zero or undefined, but even more importantly, it must actually change sign at these points.

• Find critical points by setting the second derivative, \( y'' \), to zero.
• Determine if there's a sign change just before and after each critical point.

In this problem, we calculated \( y'' = \frac{-10x^3 - 6x^2 + 6x + 2}{(1+x^2)^3} \).
By solving the equation \(-10x^3 - 6x^2 + 6x + 2 = 0 \), we found the x-values of inflection points: \( x = -1, \frac{1}{2}, \frac{4}{5} \).
To confirm these are truly points of inflection, we verified the change of sign in \( y'' \) around each \( x \) value.

Second Derivative

The second derivative of a function provides essential insights into the behavior of a curve.
While the first derivative, \( y' \), gives information about the slope and whether the function is increasing or decreasing, the second derivative, \( y'' \), indicates how the rate of change itself is changing, i.e., acceleration or concavity.
In the context of finding inflection points for \( y = \frac{1+x}{1+x^2} \), taking the second derivative involved another application of the Quotient Rule.

• This yielded \( y'' = \frac{-10x^3 - 6x^2 + 6x + 2}{(1+x^2)^3} \).

To find potential inflection points, we set \( y'' = 0 \) and solved the cubic equation \(-10x^3 - 6x^2 + 6x + 2 = 0 \).
The roots provided the critical x-values of \( -1, \frac{1}{2}, \frac{4}{5} \), where inflection points might occur, underlining the curvature of the function changes at these points.