Problem 2 - University Calculus: Early Transcendentals

Question

a. Find \(f^{-1}(x)\). b. Graph \(f\) and \(f^{-1}\) together. c. Evaluate \(d f / d x\) at \(x=a\) and \(d f^{-1} / d x\) at \(x=f(a)\) to show that at these points \(d f^{-1} / d x=1 /(d f / d x)\). $$f(x)=(1 / 5) x+7, \quad a=-1$$

Step-by-Step Solution

Answer

a. The inverse is \(f^{-1}(x)=5x-35\). b. Both graphs are symmetric about \(y=x\). c. \(d f^{-1}/d x=\frac{1}{(d f/d x)}\).

Step 1: Find the Inverse Function

To find the inverse of the function \( f(x) = \frac{1}{5}x + 7 \), we need to interchange the roles of \( x \) and \( y \) and solve for \( y \). The function first is: \( y = \frac{1}{5}x + 7 \). Swap \( x \) and \( y \) to get \( x = \frac{1}{5}y + 7 \). Then solve for \( y \): \( x - 7 = \frac{1}{5}y \). Multiply both sides by 5 to obtain \( 5(x - 7) = y \). So the inverse function is \( f^{-1}(x) = 5x - 35 \).

Step 2: Graph Both Functions

To graph \( f(x) = \frac{1}{5}x + 7 \) and \( f^{-1}(x) = 5x - 35 \), plot each function on a coordinate plane. The function \( f(x) \) is a straight line with a slope of \( \frac{1}{5} \) and y-intercept at 7. The inverse function, \( f^{-1}(x) \), is a straight line with a slope of 5 and y-intercept at -35. The graphs of a function and its inverse are symmetric with respect to the line \( y = x \).

Step 3: Differentiate f(x) and Evaluate at x = a

To find the derivative \( \frac{df}{dx} \) for \( f(x) = \frac{1}{5}x + 7 \), note that \( \frac{df}{dx} = \frac{1}{5} \) since it is a linear function. Evaluate this at \( x = a = -1 \): \( \frac{df}{dx} = \frac{1}{5} \).

Step 4: Differentiate f^{-1}(x) and Evaluate at x = f(a)

To find \( \frac{d f^{-1}}{dx} \) for \( f^{-1}(x) = 5x - 35 \), note that \( \frac{d f^{-1}}{dx} = 5 \). Evaluate this at \( x = f(-1) = \frac{1}{5}(-1) + 7 = \frac{34}{5} \). The derivative of \( f^{-1}(x) \) at this point \( x \) is \( 5 \).

Step 5: Verify the Relationship of the Derivatives

According to the inverse function theorem, \( \frac{d f^{-1}}{dx} = \frac{1}{\frac{d f}{dx}} \) at the respective points, i.e., \( \frac{d f^{-1}(x)}{dx} = \frac{1}{\frac{1}{5}} = 5 \), which is consistent with the derivative \( \frac{d f^{-1}}{dx} \), verifying the relationship \( \frac{d f^{-1}}{dx} = \frac{1}{\frac{d f}{dx}} \).

Key Concepts

Calculus

Calculus is a branch of mathematics that explores change and motion. It's a fundamental tool for understanding the world around us, especially when it comes to functions and their changes. A key concept in calculus is the idea of inverse functions. These functions essentially reverse the effect of another function. Consider a function like \( f(x) = \frac{1}{5}x + 7 \). Its inverse \( f^{-1}(x) \) is designed to undo the process of \( f(x) \). For example, if \( f(x) \) turns \( x \) into another number, \( f^{-1}(x) \) would revert it back to the original \( x \). It's like having a lock and its specific key. In the exercise above, the goal is to find the inverse of \( f(x) \), which switches the roles of \( x \) and \( y \) in the equation. Solving it, we find that \( f^{-1}(x) = 5x - 35 \). This inverse completes the key-lock analogy, making it crucial in calculus for finding solutions to equations.

Derivatives

Derivatives measure how a function changes as its input changes. Think of it as finding the rate of change or the slope of the function at a given point. In the linear function \( f(x) = \frac{1}{5}x + 7 \), the derivative \( \frac{df}{dx} \) is simply \( \frac{1}{5} \). This means that for every unit increase in \( x \), \( f(x) \) increases by \( \frac{1}{5} \).
Now, when it comes to inverse functions, there's an interesting relationship to note. According to the inverse function theorem, the derivative of an inverse function at a point is the reciprocal of the derivative of the original function at its corresponding point.
In this exercise, we see it as \( \frac{d f^{-1}}{dx} = \frac{1}{\frac{df}{dx}} \). Evaluating this at \( x = a = -1 \), we find \( \frac{df}{dx} = \frac{1}{5} \) and \( \frac{d f^{-1}}{dx} = 5 \) at \( x = \frac{34}{5} \). Thus, they validate our inverse derivative relationship perfectly.

Graphing Functions

Graphing functions provides a visual representation of how functions behave. For our function \( f(x) = \frac{1}{5}x + 7 \) and its inverse \( f^{-1}(x) = 5x - 35 \), graphing can reveal their symmetries and relationships.
These two are straight lines, each defined by their slope and y-intercept. The slope \( \frac{1}{5} \) for \( f(x) \) indicates a gentle incline, whereas the steeper slope \( 5 \) for \( f^{-1}(x) \) suggests a steeper rise.
A significant feature of functions and their inverses is their symmetry about the line \( y = x \). If you plot \( f(x) \) and \( f^{-1}(x) \) on the same graph, you'll notice that reflecting one over \( y = x \) gives you the other. This symmetry is a helpful confirmation that your inverse is correctly calculated. Graphical visualization makes these abstract functions tangible, enhancing comprehension and demonstrating important calculus properties.