Problem 38 - 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} $$f(x)=2 x^{2}, \quad x \geq 0, \quad a=5$$

Step-by-Step Solution

Answer

The inverse is \( f^{-1}(x) = \sqrt{\frac{x}{2}} \). Derivatives confirm the relationship at specified points.

Step 1: Find the Inverse Function

To find the inverse of \( f(x) = 2x^2 \), start by replacing \( f(x) \) with \( y \):\[ y = 2x^2 \]Swap \( x \) and \( y \):\[ x = 2y^2 \]Solve for \( y \): divide by 2, \[ y^2 = \frac{x}{2} \]take the square root, \[ y = \sqrt{\frac{x}{2}} \]. Thus, the inverse function is \( f^{-1}(x) = \sqrt{\frac{x}{2}} \), for \( x \geq 0 \).

Step 2: Graph f(x) and f^(-1)(x)

Graph \( f(x) = 2x^2 \), which is a parabola opening upwards, starting at \( (0,0) \) given \( x \geq 0 \).Graph \( f^{-1}(x) = \sqrt{\frac{x}{2}} \), starting at \( (0,0) \) as well, which is the upper half of a sideways parabola.Both graphs should mirror each other across the line \( y = x \).

Step 3: Differentiate f(x) and f^(-1)(x)

Differentiate \( f(x) = 2x^2 \) to get \( \frac{df}{dx} = 4x \).Differentiate \( f^{-1}(x) = \sqrt{\frac{x}{2}} \) using the chain rule or power rule. First express it as \( y = (\frac{x}{2})^{1/2} \), then differentiate to get \( \frac{d}{dx} [f^{-1}(x)] = \frac{1}{2} (\frac{x}{2})^{-1/2} \cdot \frac{1}{2} = \frac{1}{2\sqrt{2x}} \).

Step 4: Evaluate Derivatives at Specific Points

Evaluate \( \frac{df}{dx} \) at \( x=a=5 \): \( \frac{df}{dx} = 4 \times 5 = 20 \).Evaluate \( \frac{d}{dx}[f^{-1}(x)] \) at \( x=f(a)=2 \times 5^2=50 \): \( \frac{1}{2\sqrt{2\times 50}} = \frac{1}{20} \).

Step 5: Verify the Derivative Relationship

Calculate \( \frac{1}{df/dx} \) at \( x=a=5 \) to compare with \( df^{-1}/dx \) at \( x=f(a) \).\( \frac{1}{\frac{df}{dx}} = \frac{1}{20} \), which matches \( \frac{d}{dx}[f^{-1}(x)] = \frac{1}{20} \).Thus, it verifies that \( \frac{df^{-1}}{dx} = \frac{1}{\frac{df}{dx}} \) at these points.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that allows us to determine how a function changes as its input changes. Differentiating a function gives us its derivative, which represents the rate of change or slope of the function's graph at any given point.Differentiation is particularly useful for understanding and analyzing the behavior of functions, especially when it comes to optimizing or solving real-world problems. In the exercise, we begin by differentiating the function, which in this case is given as \( f(x) = 2x^2 \). To differentiate this function, we use the power rule, which states that if \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \). Applying the power rule, the derivative of \( f(x) \) becomes \( \frac{df}{dx} = 4x \). This derivative tells us that for each unit change in \( x \), the function's value changes by \( 4x \) units.To further explore these changes, we evaluate this derivative at a specific point, \( x = 5 \), yielding a change rate of 20 units at this point.This insight into the function's behavior helps us understand how fast or slow \( f(x) \) is increasing or decreasing.

Chain Rule

The chain rule is a key method in calculus for differentiating composite functions, i.e., functions within functions. When a function is composed of multiple inner and outer functions, the chain rule offers a way to differentiate it by dealing with each function step by step.The chain rule states that if a function \( h(x) = f(g(x)) \), then its derivative can be found by multiplying the derivative of \( f \) evaluated at \( g(x) \) by the derivative of \( g \) evaluated at \( x \).In our exercise, while differentiating the inverse function \( f^{-1}(x) = \sqrt{\frac{x}{2}} \), the chain rule can be applied.First, represent the inverse function as \( y = (\frac{x}{2})^{1/2} \). Notice it's a composite of the square root function and the linear function \( \frac{x}{2} \).To differentiate, follow these steps:

• Differentiating the outer function \( (g(x))^{1/2} \) gives \( \frac{1}{2}(g(x))^{-1/2} \).
• Differentiating the inner function \( g(x) = \frac{x}{2} \) gives \( \frac{1}{2} \).

Multiply the results of these derivatives to find the derivative of the entire composite function,resulting in \( \frac{1}{2\sqrt{2x}} \).This process exemplifies how the chain rule facilitates the differentiation of complex function forms.

Graphing Functions

Graphing functions is a powerful way to visualize and understand the behavior of mathematical relationships. In the exercise, we are tasked with graphing both the original function \( f(x) = 2x^2 \) and its inverse \( f^{-1}(x) = \sqrt{\frac{x}{2}} \).The graph of \( f(x) = 2x^2 \) is a parabola that opens upwards, starting at the origin (0,0) since the given domain for \( x \) is non-negative. The inverse function \( f^{-1}(x) \), on the other hand, starts at the origin as well and represents the upper half of a sideways parabola.It's important to remember that an inverse function essentially "undoes" the original function.When you plot a function and its inverse on the same graph, they reflect across the line \( y = x \). This line acts as a mirror.

• The parabola of \( f(x) \) symmetrically mirrors the sideways parabola of \( f^{-1}(x) \).
• This property helps verify if the inverse function has been derived correctly.

Graphing illustrates how the outputs of one function become the inputs of its inverse, offering both algebraic and geometric insight into their relationship.

Inverse Relationships

Inverse relationships reveal a unique aspect of functions where one function reverses or "undoes" the action of another. Finding inverse functions involves reversing the roles of the dependent and independent variables.In the exercise, we determine the inverse of \( f(x) = 2x^2 \) by solving for \( y \) in terms of \( x \), leading to the expression \( f^{-1}(x) = \sqrt{\frac{x}{2}} \).When dealing with inverse functions, an essential aspect is the relationship between their derivatives.For most well-behaved functions, the derivative of an inverse function \( \frac{df^{-1}}{dx} \) at a point is the reciprocal of the derivative of the original function \( \frac{df}{dx} \) at the corresponding point.In simpler terms:

• If \( \frac{df}{dx} \) is calculated at point \( x = a \) yielding a derivative value,
• Then the derivative \( \frac{df^{-1}}{dx} \) at \( f(a) \) should be the reciprocal of that value.

This relationship is demonstrated in the exercise where at \( x = 5 \), \( \frac{df}{dx} \) results in 20.For its inverse at \( x = f(a) = 50 \), the derivative is \( \frac{1}{20} \), confirming the reciprocal relationship exists.Such an interplay is crucial to understand not only theoretically but also practically, as it establishes a clear connection between functions and their inverses.