Problem 65 - 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. $$y=x^{3}-3 x^{2}-1, \quad 2 \leq x \leq 5, \quad x_{0}=\frac{27}{10}$$

Step-by-Step Solution

Answer

f is one-to-one; g is its inverse. Tangents at \(x_0=2.7\) are reflections across \(y=x\).

Step 1: Plot the Function and its Derivative

First, plot the function \(y = f(x) = x^3 - 3x^2 - 1\) and its derivative over the interval \(2 \leq x \leq 5\). The derivative is \(f'(x) = 3x^2 - 6x\). The function \(f\) is one-to-one if its derivative does not change sign over the interval, indicating that \(f\) is either always increasing or always decreasing. Over the interval from \(x=2\) to \(x=5\), \(f'(x)\) simplifies to \(3(x)(x-2)\), which is positive for \(x > 2\), confirming that \(f\) is one-to-one.

Step 2: Solve for the Inverse Function

Solve the equation \(y = x^3 - 3x^2 - 1\) for \(x\) in terms of \(y\), which gives the inverse function \(g(y)\). This involves rearranging the equation to express \(x\) as a function of \(y\), though finding an exact conventional expression for \(g(y)\) may require numerical methods or approximation in a typical classroom setting.

Step 3: Find the Tangent Line to f at x_0

Find the tangent line to \(f\) at the point \(x_0 = \frac{27}{10}\). First, calculate \(f'(x_0)\) to find the slope of the tangent line at this point, \(x_0 = 2.7\). Calculate \(f'(2.7)\) which is \(3(2.7)^2 - 6(2.7) = 21.87 - 16.2 = 5.67\). The equation for the tangent line at this point is given by \(y - f(2.7) = 5.67(x - 2.7)\), where \(f(2.7)\) needs to be calculated to find the full equation.

Step 4: Find the Tangent Line to g

Find the tangent line to \(g(y)\) at the corresponding point \((f(x_0), x_0)\) on the inverse. Use Theorem 1, which relates to derivative of the inverse function: If \(g = f^{-1}\), then \(g'(y) = \frac{1}{f'(g(y))}\). Compute \(g'(f(x_0)) = \frac{1}{f'(x_0)} = \frac{1}{5.67}\). The equation of the tangent at \((f(2.7), 2.7)\) becomes \(x - 2.7 = \frac{1}{5.67}(y - f(2.7))\).

Step 5: Plot and Analyze Symmetries

Plot the functions \(f\) and \(g\), the identity line \(y = x\), the two tangent lines, and the line segment joining the points \((x_0, f(x_0))\) and \((f(x_0), x_0)\). Analyze the symmetries across the main diagonal, which is the line \(y = x\). The plots should show that \(f\) and \(g\) are mirror images across the line \(y = x\), and their tangents are also reflections, highlighting the geometric symmetry inherent in inverse functions.

Key Concepts

Understanding Derivatives

Derivatives are a fundamental concept in calculus that measure how a function changes as its input changes. They're like a mathematical way of asking the question, "How fast is this function growing or shrinking?" In our exercise, the derivative of the function \( y = x^3 - 3x^2 - 1 \) is given by \( f'(x) = 3x^2 - 6x \). This derivative tells us the slope of the tangent line to the function at any given point.

Why does this matter? Because the slope essentially indicates how steep the line is at that particular point on the function graph. If the derivative is positive, it means the function is rising. If it’s negative, the function is falling. If the derivative is always positive or always negative over a specified interval, it can indicate that the function is one-to-one in that interval.

Derivatives are super useful because they help us analyze and optimize real-world problems, like finding the maximum or minimum values of a function or understanding the behavior of a function in physics or economics.

Tangent Lines and Their Significance

Tangent lines are straight lines that just barely touch a curve at a single point, without crossing it at that point. They are important because they provide a "best linear approximation" of the curve at that specific point. In simpler terms, the tangent line shows us what the curve would look like if it were a straight line right at that point.

In our exercise, the tangent line for the function \(y = x^3 - 3x^2 - 1\) at the point \(x_0 = 2.7\) is calculated using the derivative. With the slope found by \(f'(2.7) = 5.67\), the tangent line equation becomes \(y - f(2.7) = 5.67(x - 2.7)\).

This calculation is crucial because tangent lines are used to estimate values of a function near the point of tangency. They're super handy in fields such as physics, engineering, and any other discipline where you need to make predictions about values a function might take in a neighborhood of a known point.

Function Graphs and Symmetry

Graphs are visual representations of mathematical functions. They are powerful tools that allow us to see patterns and relationships in data. When looking at function graphs, important features include symmetries, intercepts, and intervals of increase or decrease. In our given exercise, the graph of the function \( y = x^3 - 3x^2 - 1 \) reveals its one-to-one nature in the interval \(2 \leq x \leq 5\) because its derivative does not change signs.

Symmetry is particularly interesting when studying a function and its inverse. A function \( f \) and its inverse \( g \) are mirrored, or symmetric, over the line \( y = x \). This means that every point \((a, b)\) on \( f \) corresponds to a point \((b, a)\) on \( g \). This symmetry is beautifully illustrated in the exercise when you plot the tangent lines and see their reflection across the line \( y=x \), which serves as the identity line where every point is equal along the x and y.

• Visual patterns help identify key points and intervals.
• Symmetry aids in understanding geometric relationships.

Applications in Calculus

Calculus powerfully enhances our understanding and ability to solve real-world problems via its principles and tools such as derivatives and tangent lines. Applications are abundant and cover numerous disciplines beyond mathematics.

For example, in physics, derivatives allow us to calculate velocities and accelerations, making it possible to predict and model the motion of objects. In economics, derivatives help optimize functions for cost and revenue, paving the way for better financial decisions.

Furthermore, understanding the symmetry of inverse functions, as shown in the exercise, finds its application in algorithms, data science, and various technological fields. Geometric insights gleaned from calculus are also integral in computer graphics and design, where understanding the nature of curves and surfaces is crucial.

In a nutshell, calculus provides a toolkit for problem-solving by breaking down complex problems into manageable calculations, helping us comprehend and influence the world more effectively.