Problem 63 - Thomas Calculus in SI Units

Question

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: a. Plot the function \(y=f(x)\) together with its derivative over the given interval. Explain why you know that \(f\) is one-to-one over the interval. b. Solve the equation \(y=f(x)\) for \(x\) as a function of \(y,\) and name the resulting inverse function \(g\) c. Find the equation for the tangent line to \(f\) at the specified point \(\left(x_{0}, f\left(x_{0}\right)\right)\) d. Find the equation for the tangent line to \(g\) at the point \(\left(f\left(x_{0}\right), x_{0}\right)\) located symmetrically across the \(45^{\circ}\) line \(y=x\) (which is the graph of the identity function). Use Theorem 1 to find the slope of this tangent line. e. Plot the functions \(f\) and \(g,\) the identity, the two tangent lines, and the line segment joining the points \(\left(x_{0}, f\left(x_{0}\right)\right)\) and \(\left(f\left(x_{0}\right), x_{0}\right) .\) Discuss the symmetries you see across the main diagonal. $$y=\frac{4 x}{x^{2}+1}, \quad-1 \leq x \leq 1, \quad x_{0}=1 / 2$$

Step-by-Step Solution

Answer

Inverse function is \(g(y) = \frac{4 - \sqrt{16 - 4y^2}}{2y}\). Tangent lines intersect at the same symmetries around \(y=x\).

Step 1: Plot Function and Derivative

Let \( f(x) = \frac{4x}{x^2+1} \). To find the derivative \( f'(x) \), use the quotient rule: \[ f'(x) = \frac{d}{dx}\left(\frac{4x}{x^2+1}\right) = \frac{(4)(x^2+1) - (4x)(2x)}{(x^2+1)^2} \] Simplifying gives: \[ f'(x) = \frac{4 - 4x^2}{(x^2+1)^2} \] Plot \( f(x) \) and \( f'(x) \) over the interval \( -1 \leq x \leq 1 \). f(x) is one-to-one over this interval since \( f'(x) \leq 0 \) for \( -1 < x < 1 \), indicating \( f \) is strictly decreasing.

Step 2: Find the Inverse Function

To find the inverse function \( g \), solve \( y = \frac{4x}{x^2+1} \) for \( x \) in terms of \( y \): \[ y(x^2 + 1) = 4x \]\[ yx^2 + y = 4x \]Re-arrange to form a quadratic: \[ yx^2 - 4x + y = 0 \]Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), find \( x \): \[ x = \frac{4 \pm \sqrt{16 - 4y^2}}{2y} \]Choose the appropriate sign ensuring \( x \) is in \( -1 \leq x \leq 1 \). Here, \( g(y) = \frac{4 - \sqrt{16 - 4y^2}}{2y} \) (since it fits the interval \([-1,1]\)).

Step 3: Tangent Line to f at \(x_0\)

Find the tangent line to \( f \) at \( x_0 = \frac{1}{2} \). Compute \( f(\frac{1}{2}) \): \[ f(\frac{1}{2}) = \frac{4 \cdot \frac{1}{2}}{(\frac{1}{2})^2 + 1} = \frac{2}{1.25} = \frac{8}{5} \] The slope at this point, \( f'(\frac{1}{2}) \):\[ f'(\frac{1}{2}) = \frac{4 - 4 \cdot (\frac{1}{4})}{(\frac{1}{4})^2} = \frac{4 - 1}{(1.25)^2} = \frac{3}{1.5625} = \frac{24}{25} \] The tangent line equation is:\[ y - \frac{8}{5} = \frac{24}{25}(x - \frac{1}{2}) \] Which simplifies to: \[ y = \frac{24}{25}x + \frac{8}{5} - \frac{12}{25} \] \[ y = \frac{24}{25}x + \frac{28}{25} \]

Step 4: Tangent Line to g using Theorem 1

By Theorem 1, the slope of the tangent to \( g \) at \( (f(x_0), x_0) \) is \( \frac{1}{f'(x_0)} \).The point where the tangent is drawn is \( (\frac{8}{5}, \frac{1}{2}) \). The slope is the reciprocal of the slope at \( x_0 \): \[ \frac{1}{\frac{24}{25}} = \frac{25}{24} \] The equation for this tangent is:\[ y - \frac{1}{2} = \frac{25}{24}(x - \frac{8}{5}) \] Which simplifies to:\[ y = \frac{25}{24}x - \frac{200}{120} + \frac{6}{12} \] \[ y = \frac{25}{24}x - \frac{8}{24} + \frac{12}{24} \] \[ y = \frac{25}{24}x + \frac{1}{6} \]

Step 5: Plot Functions and Analyze Symmetry

Plot \( f(x) \), \( g(y) \), the identity line \( y=x \), the two tangent lines from Steps 3 and 4, and the line segment joining \( (\frac{1}{2}, \frac{8}{5}) \) and \( (\frac{8}{5}, \frac{1}{2}) \). Observe that \( f \) and \( g \) are mirror images over the line \( y=x \). Tangents at symmetric points also reflect over the identity line, showcasing the inverse relationship.

Key Concepts

Derivative Calculation

Calculating the derivative of a function allows us to understand how the function's output changes with respect to its input. In simpler terms, it gives us a slope or a rate of change. Recalling the exercise, our function is given by \(f(x) = \frac{4x}{x^2+1}\). To find its derivative, we use the **Quotient Rule**. The Quotient Rule is a formula used to find the derivative of functions that are in the form of a division of two functions. If you have a function \( h(x) = \frac{u(x)}{v(x)} \), the derivative \( h'(x) \) can be found using: \[h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} \] where \( u(x) \) and \( v(x) \) are functions of \(x\), and \( u'(x) \) and \( v'(x) \) are their respective derivatives. Applying this to our problem, where \( u(x) = 4x \) and \( v(x) = x^2+1 \), we substitute into the formula. This gives us: \[ f'(x) = \frac{(4)(x^2+1) - (4x)(2x)}{(x^2+1)^2} \] Upon simplifying, we get: \[ f'(x) = \frac{4 - 4x^2}{(x^2+1)^2} \] This derivative will help us understand the behavior of \(f(x)\) and can be used to discuss its monotonicity over the interval from \(-1\) to \(1\).

Tangent Lines

A **tangent line** to a function at a particular point is a straight line that just "touches" the function at that point. It has the same slope as the function at that point and provides a linear approximation to the function near that point. In simple terms, it shows us how steep the function is at that particular spot.
For the function \(f(x)\) at \(x_0 = \frac{1}{2}\), the slope of the tangent line is determined by its derivative at \(x_0\). Using \(f'(x)\): \[ f'\left(\frac{1}{2}\right) = \frac{4 - 4\left(\frac{1}{2}\right)^2}{\left(\left(\frac{1}{2}\right)^2 + 1\right)^2} = \frac{3}{1.5625} = \frac{24}{25} \] With this slope and the point \(\left(\frac{1}{2}, \frac{8}{5}\right)\), the equation of the tangent line can be written as: \[ y - \frac{8}{5} = \frac{24}{25}\left(x - \frac{1}{2}\right) \] Which simplifies to: \[ y = \frac{24}{25}x + \frac{28}{25} \] This line provides a simple linear function that matches \(f(x)\) at \(x=\frac{1}{2}\).
Finding tangent lines is crucial in calculus because it helps in analyzing functions and making approximations.

Quotient Rule

The **Quotient Rule** is vital when differentiating expressions involving division. In our exercise, the function \(f(x) = \frac{4x}{x^2+1}\) involves both multiplication and division. Understanding how to apply the Quotient Rule here is essential.
To recap, the Quotient Rule helps us differentiate a quotient of two functions, given the basic formula: \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \] where \(u\) and \(v\) are both functions of \(x\).Applying the rule to \(f(x)\), we identify \(u(x) = 4x\) and \(v(x) = x^2 + 1\). Therefore, their derivatives are: - \(u'(x) = 4\) - \(v'(x) = 2x\)Substituting into the Quotient Rule gives us: \[ f'(x) = \frac{ (4)(x^2 + 1) - (4x)(2x) }{(x^2 + 1)^2} \] After simplification, we get: \[ f'(x) = \frac{4 - 4x^2}{(x^2 + 1)^2} \] This outcome not only reveals the slope of \(f(x)\) but underscores the importance of the Quotient Rule in calculus for handling division of functions effectively.

Function Symmetry

**Function Symmetry** is a fascinating feature to explore. Symmetric functions can be visually reflected over an axis or a line, giving us insights into their behavior. In our context, examining symmetry involves both the function \(f(x)\) and its inverse \(g(y)\).
If a function and its inverse are symmetric with respect to the line \(y=x\), their graphs reflect over this line. Let us break it down:- The interaction between \(f\) and \(g\) reflects across the line \(y = x\).- Given \(f(x) = \frac{4x}{x^2+1}\) and its inverse \(g(y)\), their symmetry across \(y = x\) confirms that \(f\) and \(g\) are truly inverses.When plotting, the graphs of \(f(x)\) and its inverse \(g(y)\), along with associated tangent lines and specific points, should also demonstrate this symmetry.This visual symmetry is not just an aesthetic experience but a conceptual confirmation of the inverse relationship. It helps cement comprehension about how functions and inverses relate and reflect in real-world or mathematical contexts.