Problem 61 - 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. $$y=\sqrt{3 x-2}, \quad \frac{2}{3} \leq x \leq 4, \quad x_{0}=3$$

Step-by-Step Solution

Answer

The function is one-to-one and increasing. Inverse: \( g(y) = \frac{y^2 + 2}{3} \). Tangent lines show symmetry across \( y = x \).

Step 1: Plot the Function and its Derivative

First, plot the function \( y = \sqrt{3x-2} \) over the interval \( \frac{2}{3} \leq x \leq 4 \). Calculate the derivative: \( y' = \frac{3}{2\sqrt{3x-2}} \). Verify that \( y' > 0 \) for all \( x \) in the interval, indicating that \( f \) is strictly increasing and thus one-to-one.

Step 2: Find the Inverse of the Function

To find the inverse, solve \( y = \sqrt{3x-2} \) for \( x \). Square both sides to get \( y^2 = 3x - 2 \), then solve for \( x \) to get \( x = \frac{y^2 + 2}{3} \). Therefore, the inverse function is \( g(y) = \frac{y^2 + 2}{3} \).

Step 3: Find the Tangent Line to f at (x_0, f(x_0))

Evaluate the function and its derivative at \( x_0 = 3 \). The function value is \( f(3) = \sqrt{7} \), and the derivative is \( f'(3) = \frac{3}{2\sqrt{7}} \). The equation of the tangent line is \( y = \frac{3}{2\sqrt{7}}(x - 3) + \sqrt{7} \).

Step 4: Find the Tangent Line to g at (f(x_0), x_0)

First, find the slope of the tangent to \( g \) using Theorem 1: if \( y = f(x) \) implies \( x = g(y) \), then \( g'(y) = \frac{1}{f'(g(y))} \). So, at \( y_0 = \sqrt{7} \), \( g at y_0 = 3 \), \( g' = \frac{2\sqrt{7}}{3} \). The tangent line is \( y = \frac{2\sqrt{7}}{3}(x - \sqrt{7}) + 3 \).

Step 5: Plot and Discuss Symmetries

Plot \( y = \sqrt{3x-2} \), \( x = \frac{y^2 + 2}{3} \), the identity line \( y = x \), both tangent lines, and the line connecting \( (3, \sqrt{7}) \) and \( (\sqrt{7}, 3) \). Notice the symmetry of the graphs and tangent lines across the line \( y = x \), highlighting the relationship between functions and their inverses.

Key Concepts

Derivative

When examining functions and their behavior, derivatives are a powerful tool. The derivative of a function, often denoted as \( f'(x) \), gives us the rate at which the function's output changes with respect to a change in the input. Think of it as the slope of a curve at any point. For a function \( y = \sqrt{3x - 2} \), its derivative is \( y' = \frac{3}{2\sqrt{3x-2}} \). To better understand this, remember that the derivative can tell us if a function is increasing or decreasing.

• If \( y' > 0 \), the function is increasing.
• If \( y' < 0 \), the function is decreasing.

For our exercise, since \( y' > 0 \) within the interval \( \frac{2}{3} \leq x \leq 4 \), the function is strictly increasing here, which is a key factor in determining its one-to-one nature.

One-to-One Function

A function is called one-to-one if it never takes the same value twice — every output has exactly one input. This characteristic is often checked using the "Horizontal Line Test". If no horizontal line crosses the graph more than once, the function is one-to-one.To ensure our function \( y = \sqrt{3x-2} \) is one-to-one over the specified interval, we rely on the nature of derivatives. Since the derivative \( y' \) is always positive over the interval \( \frac{2}{3} \leq x \leq 4 \), it means the function is always increasing. Hence, it will not repeat any values, confirming its one-to-one property.

Tangent Line

The tangent line to a curve at a given point provides an approximation of the curve at that point. It's like zooming in so closely that the curve appears straight. The equation of the tangent line is derived from the point-slope form of a line: \[ y - f(x_0) = f'(x_0)(x - x_0) \]For our function \( f(x) = \sqrt{3x-2} \) at \( x_0 = 3 \), the tangent line can be calculated as:

• At \( x=3 \), \( f(3) = \sqrt{7} \)
• The slope is \( f'(3) = \frac{3}{2\sqrt{7}} \)

Thus, the tangent line equation is \( y = \frac{3}{2\sqrt{7}}(x - 3) + \sqrt{7} \). This line closely follows the curve at the point \( (3, \sqrt{7}) \).

Symmetry

Symmetry in functions often provides insights into their relationships and properties. In our context, the symmetry across the line \( y = x \) is particularly important, as it characterizes the relationship between a function and its inverse.When we plot both \( y = f(x) \) and its inverse \( x = g(y) \), their graphs appear as mirror images across the line \( y = x \). This line is known as the identity line and acts as an axis of symmetry.

• The tangent lines at points \((x_0, f(x_0))\) on the original graph and \((f(x_0), x_0)\) on the inverse graph also reflect this symmetry.
• The line joining points \((3, \sqrt{7})\) and \((\sqrt{7}, 3)\) further exemplifies this mirrored relationship.

Recognizing these symmetrical patterns helps to visually confirm the theoretical calculations about inverses and derivatives, making it easier to understand their interconnections.