Problem 37 - Thomas Calculus

Question

In Exercises \(35-38 :\) \begin{equation} \begin{array}{l}{\text { a. Find } f^{-1}(x) \text { . }} \\ {\text { b. Graph } f \text { and } f^{-1} \text { together. }} \\ {\text { c. Evaluate } d f / d x \text { at } x=a \text { and } d f^{-1} / d x \text { at } x=f(a) \text { to show }} \\ {\text { that at these points } d f^{-1} / d x=1 /(d f / d x)}\end{array} \end{equation} $$f(x)=5-4 x, \quad a=1 / 2$$

Step-by-Step Solution

Answer

The inverse function is \( f^{-1}(x) = \frac{5-x}{4} \). At \( x = \frac{1}{2} \), \( \frac{df}{dx} = -4 \) and \( \frac{df^{-1}}{dx} = -\frac{1}{4} \) at \( x = 3 \), proving the reciprocal relationship.

Step 1: Find the Inverse Function

To find the inverse function of \( f(x) = 5 - 4x \), we first replace \( f(x) \) with \( y \) for convenience. Thus, \( y = 5 - 4x \). To find the inverse, we swap \( x \) and \( y \) and solve for \( y \). Swap \( x \) and \( y \) to get \( x = 5 - 4y \). Solve for \( y \):\[ 4y = 5 - x \] \[ y = \frac{5-x}{4} \]. The inverse function is \( f^{-1}(x) = \frac{5-x}{4} \).

Step 2: Graph f and f^{-1} Together

To graph \( f(x) = 5 - 4x \) and \( f^{-1}(x) = \frac{5-x}{4} \) together, plot both equations on the same set of axes. The function \( f(x) \) is a straight line with a slope of \(-4\) and a y-intercept of 5. The inverse \( f^{-1}(x) \) is also a straight line but with a slope of \(-\frac{1}{4}\) and a y-intercept of \( \frac{5}{4} \). These lines should reflect across the line \( y = x \), which is the characteristic of a function and its inverse.

Step 3: Differentiate f and Evaluate at x=a

Differentiate \( f(x) = 5 - 4x \) with respect to \( x \): \( \frac{df}{dx} = -4 \). Evaluate this derivative at \( x = a = \frac{1}{2} \): since \( \frac{df}{dx} = -4 \) is constant, it remains \( -4 \) for any value of \( x \).

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

Find \( f(a) \) where \( a = \frac{1}{2} \). \( f(\frac{1}{2}) = 5 - 4(\frac{1}{2}) = 3 \). Now, differentiate \( f^{-1}(x) = \frac{5-x}{4} \): \( \frac{d}{dx} \left( \frac{5-x}{4} \right) = -\frac{1}{4} \). Evaluate this derivative at \( x = f(a) = 3 \), which remains \( -\frac{1}{4} \).

Step 5: Show Relationship Between Derivatives

Verify that \( \frac{d f^{-1}}{dx} \) is the reciprocal of \( \frac{df}{dx} \). At \( x = a \), \( \frac{df}{dx} = -4 \); at \( x = f(a) = 3 \), \( \frac{d f^{-1}}{dx} = -\frac{1}{4} \). The relationship \( \frac{d f^{-1}}{dx} = \frac{1}{\frac{df}{dx}} \) holds: \( -\frac{1}{4} = \frac{1}{-4} \).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the rate at which a function changes. It's like discovering the function's speed at different points along its curve. For any given function, differentiation provides its derivative, symbolized by \( \frac{df}{dx} \), which represents the function's slope or steepness at any point.
To differentiate a linear function like \( f(x) = 5 - 4x \), we simply look at the coefficient of \( x \), which is \(-4\). Therefore, the derivative is a constant \(-4\), meaning the function decreases at a consistent rate as x increases. This tells us that for every unit increase in \( x \), \( f(x) \) decreases by 4 units.
Once the derivative is determined, we can evaluate it at any specific point, such as \( x = 1/2 \), to see the rate of change exactly there. With linear functions, this will be the same everywhere on the line.

Graphing Functions

Graphing functions helps visualize how a function behaves across different values of \( x \). When dealing with a function and its inverse, like \( f(x) = 5 - 4x \) and its inverse \( f^{-1}(x) = \frac{5-x}{4} \), we gain insights by plotting them together.
Both functions are straight lines, but they intersect and reflect over the line \( y = x \), showcasing a classic property of inverse functions. For \( f(x) \), with its negative slope of \(-4\), the line descends steeply, while \( f^{-1}(x) \) has a gentler downward slope of \(-\frac{1}{4}\). This visual symmetry across \( y = x \) is an essential feature in understanding how original and inverse functions relate to one another on a graph.

• Identify slopes and y-intercepts for setup.
• Sketch primary functions and their inverses.
• Observe line \( y = x \) for symmetry between function pairs.

Reciprocal Derivatives

When dealing with inverse functions, a fascinating relationship emerges between their derivatives: the derivative of an inverse function is the reciprocal of the original's derivative. When you differentiate a function like \( f(x) = 5 - 4x \), you get \( \frac{df}{dx} = -4 \). Similarly, for the inverse function \( f^{-1}(x) = \frac{5-x}{4} \), differentiation yields \( \frac{d f^{-1}}{dx} = -\frac{1}{4} \).
Evaluating these derivatives at specific points can be incredibly illuminating. At point \( x = a = 1/2 \), the derivative of \( f(x) \) remains constant at \(-4\). Don't forget, the reciprocal relationship holds here. So when you evaluate \( f^{-1} \) at the corresponding point \( x = f(a) = 3 \), its derivative is \(-\frac{1}{4}\), perfectly being the reciprocal of \(-4\) (i.e., \( \frac{1}{-4} \)).
This interplay of derivatives is a powerful tool for verifying inverse relationships and understanding how inverse functions respond inversely in terms of their rates of change.