Problem 53 - Thomas Calculus

Question

In Exercises \(53-60,\) 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 within the interval. Its inverse is \(g(y) = \frac{y^2 + 2}{3}\). Tangent lines were derived for both the function and its inverse, demonstrating symmetry across the identity line \(y = x\).

Step 1: Plot the Function and Its Derivative

For the function \(y = \sqrt{3x - 2}\), we first calculate its derivative using the chain rule. The derivative \(f'(x)\) is given by: \[ f'(x) = \frac{d}{dx}(\sqrt{3x - 2}) = \frac{3}{2\sqrt{3x - 2}}. \]To confirm that \(f\) is one-to-one over \(\frac{2}{3} \leq x \leq 4\), observe that \(f'(x) > 0\) for all \(x\) in this interval, indicating that \(f\) is increasing and thus one-to-one.

Step 2: Solve for Inverse Function

Solve \(y = \sqrt{3x - 2}\) for \(x\) in terms of \(y\). Squaring both sides gives \(y^2 = 3x - 2\), which simplifies to \[ 3x = y^2 + 2. \]Thus, solving for \(x\), we find the inverse function \(g(y) = \frac{y^2 + 2}{3}.\) Hence, the inverse is \(g(y) = \frac{y^2 + 2}{3}.\)

Step 3: Find Tangent Line to Function

To find the tangent line to \(f\) at \((x_0, f(x_0)) = (3, \sqrt{3 \cdot 3 - 2}) = (3, \sqrt{7})\), use the point-slope form of a line, with \(f'(3) = \frac{3}{2\sqrt{7}}\). The tangent line is:\[y - \sqrt{7} = \frac{3}{2\sqrt{7}}(x - 3).\] Simplifying gives the equation of the tangent line.

Step 4: Find Tangent Line to Inverse Function

To find the tangent line to the inverse \(g\) at \((f(x_0), x_0)=(\sqrt{7}, 3)\), Theorem 1 states that the slope of the tangent line to \(g\) at this point is the reciprocal of \(f'(x_0)\), giving slope \(\frac{2\sqrt{7}}{3}\). Applying the point-slope form results:\[x - 3 = \frac{2\sqrt{7}}{3}(y - \sqrt{7}).\] Simplifying gives the tangent line equation.

Step 5: Plot All Relevant Functions and Lines

Plot the original function \(f(x) = \sqrt{3x - 2}\), its inverse \(g(y) = \frac{y^2 + 2}{3}\), the identity line \(y = x\), and the two tangent lines found in the previous steps. Additionally, plot the line segment joining the points \((3, \sqrt{7})\) and \((\sqrt{7}, 3)\). Note the symmetry about the line \(y = x\), as each function and its tangent reflect across this line.

Key Concepts

Derivatives

Derivatives are a fundamental concept in calculus. They represent the rate at which a function changes with respect to changes in its input. For any function, its derivative is itself a new function that gives us the slope of the tangent line to the curve at any given point.

• For the function provided, namely, \( y = \sqrt{3x - 2} \), the derivative \( f'(x) \) is \( \frac{3}{2\sqrt{3x - 2}} \).
• This derivative is positive for all \( x \) in the interval \( \frac{2}{3} \leq x \leq 4 \).
• When \( f'(x) > 0 \), it indicates that the function is increasing over that interval.

This increasing nature of the function further implies that it is one-to-one, as no horizontal line intersects the curve more than once within the given interval. This property is crucial when considering inverse functions.

Tangent Lines

Tangent lines are lines that touch a curve at only one point and have the same slope as the curve at that point. These lines are extremely helpful for approximations and for understanding the behavior of functions locally.
For the function \( f(x) = \sqrt{3x - 2} \), we can find the tangent line at a given point \( (3, \sqrt{7}) \):

• The slope of the tangent is given by the function's derivative, \( f'(3) = \frac{3}{2\sqrt{7}} \).
• Using the point-slope form, the equation becomes \( y - \sqrt{7} = \frac{3}{2\sqrt{7}}(x - 3) \).

Tangent lines to curves provide an excellent linear approximation to functions at given points and are a key part of differential calculus. Similarly, for the inverse function \( g(y) \), the tangent at the point \( (\sqrt{7}, 3) \) can be found using the reciprocal of the slope of the tangent line to \( f \).

Chain Rule

The chain rule is a method in calculus for finding the derivative of the composition of two functions. This is incredibly useful when dealing with more complicated functions and enables us to understand how changes in one function affect another.
In this scenario, we used the chain rule to find the derivative \( f'(x) \) of the function \( y = \sqrt{3x - 2} \):

• The outer function is \( y = \sqrt{u} \) with \( u = 3x - 2 \).
• The derivative of \( \sqrt{u} \) is \( \frac{1}{2\sqrt{u}} \), and the derivative of \( u \) is \( 3 \).
• Thus, the combined derivative \( f'(x) = \frac{3}{2\sqrt{3x - 2}} \).

Applying the chain rule helps us evaluate derivatives of nested functions and is a core technique in advanced calculus studies.

One-to-One Functions

A function is one-to-one if each output value corresponds to exactly one input value. This is an important property when discussing inverse functions, as only one-to-one functions have inverses that are also functions.
For the function \( f(x) = \sqrt{3x - 2} \), being one-to-one is verified by the fact that it is monotonically increasing (i.e., always increasing) on the given interval \( \frac{2}{3} \leq x \leq 4 \).

• Monotonicity—demonstrated by \( f'(x) > 0 \)—ensures that \( f(x) \) is always increasing.
• This quality confirms that \( f \) is one-to-one over the specified interval.
• This implies that each \( x \) maps to a unique \( y \), allowing for a well-defined inverse function \( g(y) \).

Understanding and determining whether a function is one-to-one is crucial for solving and finding inverse relations, essential in calculus and beyond.