Problem 63 - Thomas Calculus

Question

In Exercises \(61-66,\) you will explore some functions and their inverses together with their derivatives and linear approximating functions at specified points. Perform the following steps using your CAS: \begin{equation} \begin{array}{l}{\text { a. Plot the function } y=f(x) \text { together with its derivative over the given }} \\ {\text { interval. Explain why you know that } f \text { is one-to-one over the interval. }} \\ {\text { b. Solve the equation } y=f(x) \text { for } x \text { as a function of } y, \text { and name }} \\ {\text { the resulting inverse function } g \text { . }} \\\ {\text { c. Find the equation for the tangent line to } f \text { at the specified point }} \\ {\quad\left(x_{0}, f\left(x_{0}\right)\right) .}\\\\{\text { d. Find the equation for the tangent line to } g \text { at the point }\left(f\left(x_{0}\right), x_{0}\right)} \\ {\text { located symmetrically across the } 45^{\circ} \text { line } y=x \text { (which is the }} \\ {\text { graph of the identity function). Use Theorem } 1 \text { to find the slope }} \\ {\text { of this tangent line. }}\\\\{\text { e. Plot the functions } f \text { and } g \text { , the identity, the two tangent lines, and }} \\ {\text { the line segment joining the points }\left(x_{0}, f\left(x_{0}\right)\right) \text { and }\left(f\left(x_{0}\right), x_{0}\right) .} \\ {\text { Discuss the symmetries you see across the main diagonal. }}\end{array} \end{equation} $$y=\frac{4 x}{x^{2}+1}, \quad-1 \leq x \leq 1, \quad x_{0}=1 / 2$$

Step-by-Step Solution

Answer

The function is one-to-one as its derivative sign is constant. The inverse is \( g(y) = \frac{4 - \sqrt{16 - 4y^2}}{2y} \). The tangent lines are linearly approximated and show symmetry across \( y = x \).

Step 1: Plot the Function and Derivative

Plot the function \( y = \frac{4x}{x^2 + 1} \) over the interval \(-1 \leq x \leq 1\) along with its derivative. The derivative \( f'(x) \) is calculated as \( \frac{d}{dx}\left(\frac{4x}{x^2+1}\right) \). Use the quotient rule to find \( f'(x) = \frac{4(1-x^2)}{(x^2+1)^2}\). The function is one-to-one over this interval because its derivative does not change sign, indicating the function is either strictly increasing or decreasing.

Step 2: Find the Inverse Function

Solve the equation \( y = \frac{4x}{x^2 + 1} \) for \( x \) in terms of \( y \). Rearranging gives \( y(x^2 + 1) = 4x \) which becomes \( yx^2 + y = 4x \). Rearrange into a quadratic form: \( yx^2 - 4x + y = 0 \). Use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = y \), \( b = -4 \), and \( c = y \). Solving gives the inverse \( g(y) = \frac{4 \pm \sqrt{16 - 4y^2}}{2y} \). Since the inverse is in the range \(-1 \leq x \leq 1\), use \( g(y) = \frac{4 - \sqrt{16 - 4y^2}}{2y} \).

Step 3: Find the Tangent Line to the Function

Find the equation of the tangent line to \( f(x) \) at \( x_0 = \frac{1}{2} \). First, calculate \( f\left(\frac{1}{2}\right) = \frac{4 \times \frac{1}{2}}{\frac{1}{4} + 1} = \frac{4 \times \frac{1}{2}}{\frac{5}{4}} = \frac{2}{\frac{5}{4}} = \frac{8}{5} \). The derivative \( f'(x) = \frac{4(1-x^2)}{(x^2+1)^2} \), evaluated at \( x_0 = \frac{1}{2} \) gives \( f'\left(\frac{1}{2}\right) = \frac{4(1-(\frac{1}{2})^2)}{((\frac{1}{2})^2+1)^2} = \frac{4(\frac{3}{4})}{(\frac{5}{4})^2} = \frac{12}{25} \). The tangent line is \( y - \frac{8}{5} = \frac{12}{25}(x - \frac{1}{2}) \).

Step 4: Find the Tangent Line to the Inverse

The point on the inverse function at \((f\left(x_0\right), x_0)\) is \((\frac{8}{5}, \frac{1}{2})\). Using Theorem 1, the slope of the tangent line to the inverse \( g(y) \) at this point is the reciprocal of the slope of the tangent line to \( f(x) \), which is \( \frac{25}{12} \). Therefore, the equation for this tangent line is \( x - \frac{1}{2} = \frac{25}{12}(y - \frac{8}{5}) \).

Step 5: Plot Everything and Analyze Symmetry

Plot the functions \( f(x) \) and \( g(y) \), their tangent lines, the identity function \( y = x \), and the line segment joining \((x_0, f(x_0))\) to \((f(x_0), x_0)\). Observe the symmetry about the line \( y = x \), noting how the tangent lines reflect across this line.

Key Concepts

one-to-one functions

A function is considered one-to-one if it never assigns the same value to two different inputs. This means that for any two different values of

• x
• the outputs (or y-values) must also be different

Markedly, one-to-one functions have a special feature—they possess unique inverses. To confirm the function given in the exercise is one-to-one, we can look at its derivative.
If the derivative does not change sign over the given interval,

• it indicates the function does not change from increasing to decreasing, or vice versa.

This is the case for the function defined by y = \( \frac{4x}{x^2 + 1} \) within the interval \(-1 \leq x \leq 1\), ensuring it's strictly one-to-one over this region.

derivatives

Derivatives are a fundamental tool in calculus. They tell us how a function changes at any given point. In simpler terms, a derivative gives us the slope of the tangent line to the function at a particular point.
To find the derivative of the exercise's function y = \( \frac{4x}{x^2 + 1}\), we employ the quotient rule.

• The quotient rule is used because the function is a ratio of two functions.

The rule states that if \( u(x) \) and \( v(x) \) are differentiable functions, then the derivative of their ratio

• \( \frac{u}{v} \)
• is \( \frac{u'v - uv'}{v^2} \)

Combining this with our function leads us to the derivative: \( f'(x) = \frac{4(1-x^2)}{(x^2+1)^2} \). This derivative helps in confirming the one-to-oneness and also aids in finding tangent lines later.

tangent lines

Tangent lines are straight lines that just "touch" a curve at a particular point and have the same slope as the curve at that point. For the function given in the exercise, the tangent line at

• \( x = \frac{1}{2} \)

acts as an approximation of the function near this point.
Knowing the slope of the tangent, which is simply the derivative of the function, allows us to easily write its equation.
We use the point-slope form of a line:

• \( y - y_1 = m(x - x_1) \)

Here, \( m \) is the slope derived from \( f'(x) \), and

• \( (x_1, y_1) \)
• is the point on the curve, specifically

\( \left(\frac{1}{2}, \frac{8}{5}\right) \).
The slope at this point is calculated as \( \frac{12}{25} \), leading to the tangent line equation: \( y - \frac{8}{5} = \frac{12}{25}(x - \frac{1}{2}) \).

symmetry

Symmetry in mathematics can usually indicate an inherent balance or structure in functions
In terms of graphing functions and inverses, one important type of symmetry occurs across the line

• \( y = x \)

This is especially relevant with inverse functions.
Each point on the function and its inverse is reflected across this diagonal line that acts like a mirror.

• For example, if a point
• \( (a, b) \)
• lies on the original function, the point

\( (b, a) \) perfectly reflects on the inverse.
In the exercise, plotting the tangent lines to the function and its inverse shows how they too reflect across the line

• \( y = x \)

. Observing this symmetry helps to affirm the correctness of your inverse calculations.

quadratic formula

The quadratic formula is a solution to solving quadratic equations of the form

• \( ax^2 + bx + c = 0 \)

It is given as

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

In context of the exercise, while deriving the inverse of the function y = \( \frac{4x}{x^2 + 1} \), we rearrange it into a quadratic equation format by isolating terms around

• the variable \( x \)

. Substituting the rearranged equation coefficients

• (\( a = y \), \( b = -4 \), \( c = y \))

into the quadratic formula, solves the inverse in terms of

• \( x \)

This crucial step ensures that we can express x in terms of y, bridging our work towards finding the inverse of the given function.

quotient rule

The quotient rule is central to finding the derivative of a rational function, which is essentially a function expressed as one function divided by another.
The rule states:

• If a function \( f(x) = \frac{u(x)}{v(x)} \),
• where \( u(x) \) and \( v(x) \) are both differentiable,

then its derivative is \( f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2} \)
In the exercise's problem with

• \( y = \frac{4x}{x^2 + 1} \),

we need to differentiate the numerator (4x) and the denominator (x^2 + 1),

• which yields their derivatives as
• \( 4 ext{ and } 2x \), respectively.

Using the quotient rule, these derivatives are used to arrive at \( f'(x) = \frac{4(1-x^2)}{(x^2+1)^2} \)
Understanding this rule is pivotal to comfortably dealing with complex rational functions.

calculus

Calculus is the mathematical study of continuous change. It allows us to model systems, predict changes, and understand the nature of function behaviors. Central to calculus are concepts such as derivatives and integrals.
Derivatives, such as those we calculated in this exercise, help us find rates of change (slopes) essentially "speed" at which functions change.
On the other hand, integrals, although not used specifically in this problem, would help us find areas under curves or accumulate quantities.
The connection of inverse functions and derivatives exemplifies calculus's strength in handling dynamic quantities.

• These foundations allow us to see relationships that different mathematical expressions can have

like symmetries and reflections, expanding our ability to carry out complex analyses smoothly and systematically.