Problem 28 - 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 \(d f^{-1} / d x=1 /(d f / d x) .\) $$ f(x)=2 x^{2}, \quad x \geq 0, \quad a=5 $$

Step-by-Step Solution

Answer

The function's inverse is \( f^{-1}(x) = \sqrt{\frac{x}{2}} \), and derivatives confirm \( df^{-1}/dx = 1/(df/dx)\) at given points.

Step 1: Find the inverse function

Given the function \( f(x) = 2x^2 \) for \( x \geq 0 \), we need to find its inverse \( f^{-1}(x) \). Start by setting \( y = 2x^2 \). Solve for \( x \) in terms of \( y \). This gives \( x = \sqrt{\frac{y}{2}} \). Therefore, the inverse function is \( f^{-1}(x) = \sqrt{\frac{x}{2}} \).

Step 2: Graph the functions

Plot the function \( f(x) = 2x^2 \) and its inverse \( f^{-1}(x) = \sqrt{\frac{x}{2}} \). Since the domain of \( f(x) \) is \( x \geq 0 \), its inverse will be defined for \( x \geq 0 \) as well. The graph of an inverse function is a reflection of the original function across the line \( y = x \).

Step 3: Evaluate derivatives at specific points

Calculate the derivative of \( f(x) = 2x^2 \), which is \( \frac{df}{dx} = 4x \). Evaluate this at \( x = a = 5 \), giving \( \frac{df}{dx} \bigg|_{x=5} = 20 \). Now calculate the derivative of \( f^{-1}(x) = \sqrt{\frac{x}{2}} \), which is \( \frac{d f^{-1}}{dx} = \frac{1}{2\sqrt{2x}} \). Evaluate it at \( x = f(5) = 2 \times 5^2 = 50 \), giving \( \frac{d f^{-1}}{dx} \bigg|_{x=50} = \frac{1}{20} \).

Step 4: Verify the reciprocal relationship

Since \( \frac{df^{-1}}{dx} \bigg|_{x=f(a)} = \frac{1}{20} \) and \( \frac{df}{dx} \bigg|_{x=a} = 20 \), we see that \( \frac{df^{-1}}{dx} \bigg|_{x=f(a)} = \frac{1}{\frac{df}{dx} \bigg|_{x=a}} \), confirming the reciprocal relationship \( \frac{df^{-1}}{dx} = \frac{1}{\frac{df}{dx}} \) at these points.

Key Concepts

Understanding Derivatives in Inverse Functions

Derivatives measure how a function changes as its input changes. For any given function, the derivative at a certain point can tell you how steep the tangent line there is. When dealing with inverse functions, derivatives play a crucial role.
The original function is often represented as \( f(x) \) and the inverse function as \( f^{-1}(x) \). Knowing how to calculate it, using principles from calculus, is fundamental.
Calculating derivatives involves using rules such as the power rule. For \( f(x) = 2x^2 \), the derivative is \( \frac{df}{dx} = 4x \). This gives the rate of change of \( f(x) \) along its curve. By evaluating this at a specific \( x \), we can find exactly how fast the function is changing at that point.
If the derivative of the original function is known, one can figure out the behavior of the inverse function's derivative through the reciprocal relationship. This requires more than just straightforward calculations; understanding the concept is key.

Graphing Functions and Their Inverses

Graphing a function and its inverse allows us to visualize their relationship. For any function \( f(x) \), its inverse \( f^{-1}(x) \) reflects across the line \( y = x \). The graphs mirror each other symmetrically.To illustrate, consider \( f(x) = 2x^2 \), with its inverse \( f^{-1}(x) = \sqrt{\frac{x}{2}} \). Both functions define a unique path along the x-y plane.
For \( x \geq 0 \), the graph of \( f(x) \) is a rise-up parabola, while \( f^{-1}(x) \) forms a gentler slope due to its square root nature.

• For accurate graphing, confirm both functions meet at key points where \( f(f^{-1}(x)) = x \).
• The line \( y=x \) acts as the axis for their symmetric difference.

By graphing the function and its inverse, we can better comprehend their intricate linkage.

Exploring the Reciprocal Relationship

The reciprocal relationship between a function and its inverse's derivatives is a fascinating concept. For an invertible function \( f(x) \) and its inverse \( f^{-1}(x) \), there holds the property: \( \frac{df^{-1}}{dx} = \frac{1}{\frac{df}{dx}} \). This property is particularly significant at specific points.
If you evaluate the derivative of \( f(x) = 2x^2 \) at, for instance, \( x = 5 \), it results in \( 20 \). At this point, the inverse's derivative at \( x = 50 \) yields \( \frac{1}{20} \). Here, they exhibit mathematically inverse or reciprocal behaviors.
Understanding and proving this relationship strengthens one's grasp of calculus principles.
This insight is not only theoretical: it has practical applications in fields that require the clarity of function behavior, such as physics and engineering.

• Always verify this reciprocal nature by checking each calculation step.
• Use this principle to ensure accuracy when working with complex function pairs.

By embracing this unique relationship, students find more depth in understanding how inverse functions operate.