Problem 27 - Thomas Calculus

Question

In Exercises \(25-28 :\) 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 \(d f^{-1} / d x=1 /(d f / d x) .\) $$ f(x)=5-4 x, \quad a=1 / 2 $$

Step-by-Step Solution

Answer

The inverse is \( f^{-1}(x) = \frac{5-x}{4} \), and both derivatives are consistently \( -4 \) and \( -\frac{1}{4} \).

Step 1: Find the Inverse Function

To find the inverse function \( f^{-1}(x) \), first replace \( f(x) \) with \( y \): \[ y = 5 - 4x \]Solve for \( x \): \[ 4x = 5 - y \]\[ x = \frac{5-y}{4} \]Now, replace \( y \) with \( x \) for the inverse function:\[ f^{-1}(x) = \frac{5-x}{4} \].

Step 2: Graph the Functions

To graph both \( f(x) = 5 - 4x \) and \( f^{-1}(x) = \frac{5-x}{4} \), plot each function on the same set of axes. The function \( f(x) \) is a straight line with a slope of -4, and the inverse function \( f^{-1}(x) \) is a line with a slope of -1/4. Both graphs should reflect across the line \( y = x \).

Step 3: Differentiate the Original Function

Find the derivative of \( f(x) \):\[ \frac{df}{dx} = -4 \].

Step 4: Differentiate the Inverse Function

Find the derivative of \( f^{-1}(x) \):\[ \frac{d f^{-1}}{d x} = -\frac{1}{4} \].

Step 5: Evaluate Derivatives at Specific Points

Evaluate \( \frac{df}{dx} \) at \( x = a = \frac{1}{2} \): This remains \( -4 \) because \( \frac{df}{dx} = -4 \) everywhere.For \( f^{-1} \), first find \( f(a) \):\[ f\left(\frac{1}{2}\right) = 5 - 4 \cdot \frac{1}{2} = 3 \]Now evaluate \( \frac{d f^{-1}}{d x} \) at \( x = 3 \): This remains \( -\frac{1}{4} \) because \( \frac{d f^{-1}}{d x} = -\frac{1}{4} \) everywhere.

Step 6: Verify the Derivative Relationship

Verify that \( \frac{d f^{-1}}{d x} = \frac{1}{\frac{df}{dx}} \) by checking:\[ -\frac{1}{4} = \frac{1}{-4} \]This equality confirms that the relationship holds true at the specified points.

Key Concepts

Graphing Functions

Graphing functions involves plotting their graphs on a Cartesian coordinate system, which allows us to visually examine their behavior. In the problem at hand, we have two functions: the original function \( f(x) = 5 - 4x \) and its inverse \( f^{-1}(x) = \frac{5-x}{4} \). Both functions are linear, meaning their graphs are straight lines.

• For the function \( f(x) = 5 - 4x \), the slope is \(-4\), which means it decreases steeply. The y-intercept is \(5\), so the graph crosses the y-axis at \( y = 5 \).
• The inverse function \( f^{-1}(x) = \frac{5-x}{4} \) has a slope of \(-\frac{1}{4}\). It decreases more gently, and the y-intercept is \( \frac{5}{4} \).

The key feature when graphing an inverse function is symmetry across the line \( y = x \). This line serves as a reflective line, meaning each point \((a, b)\) on \( f(x) \) will correspond to \((b, a)\) on \( f^{-1}(x) \). To accurately represent these graphs, ensure to plot each function on the same axis and look for this symmetric reflection.

Differentiation

Differentiation is a foundational concept in calculus that allows us to determine the rate at which a function changes. This rate is represented by the derivative of the function. When we differentiate a linear function such as \( f(x) = 5 - 4x \), we use standard rules of differentiation.

• The derivative of a constant is zero. Thus, differentiating the constant \(5\) yields \(0\).
• For the term \(-4x\), the derivative is only the coefficient \(-4\), following the power rule \( \frac{d}{dx}[ax] = a \).

Hence, the derivative of \( f(x) \) is constant and equal to \(-4\) everywhere.Differentiation is a vital tool for analyzing function behaviors, such as determining slopes of tangents, finding maxima and minima, and understanding instantaneous rates of change. In this problem, knowing \( \frac{df}{dx} \) helps us connect to the concept of inverse functions through their derivatives.

Derivative of Inverse Functions

Inverse functions and their derivatives exhibit a particular characteristic. The derivative of an inverse function at a point is the reciprocal of the derivative of the original function at its corresponding point. This property holds under certain conditions, primarily when both functions are differentiable, and the given inputs and outputs are sufficiently close.For the given functions, the derivative of the inverse \( f^{-1}(x) = \frac{5-x}{4} \) is found as follows:

• We differentiate \( f^{-1}(x) \), yielding \(-\frac{1}{4} \).

When evaluated at \( x = 3 \) (the output of \( f(a) \)), it remains \(-\frac{1}{4} \).The relationship \( \frac{d f^{-1}}{d x} = \frac{1}{\frac{df}{dx}} \) holds true because \(-\frac{1}{4} = \frac{1}{-4} \). Understanding this key relationship enhances comprehension of how inverse functions behave and provides an intuitive way to explore more complex systems where these functions play a role.