Problem 43 - Thomas Calculus in SI Units

Question

Suppose that the differentiable function \(y=f(x)\) has an inverse and that the graph of \(f\) passes through the point (2,4) and has a slope of \(1 / 3\) there. Find the value of \(d f^{-1} / d x\) at \(x=4\)

Step-by-Step Solution

Answer

The value of \( df^{-1}/dx \) at \( x = 4 \) is 3.

Step 1: Understanding Inverses and Derivatives

When a function has an inverse, we can use the inverse function theorem, which states that the derivative of the inverse function at a point is the reciprocal of the derivative of the original function at the corresponding point. Here we need to find \( \frac{d}{dx} f^{-1}(x) \) at \( x = 4 \).

Step 2: Identifying Given Points and Slopes

We know that the function \( f \) passes through the point \( (2,4) \) and has a slope of \( \frac{1}{3} \) at \( x=2 \). This means that for the inverse function, \( f^{-1} \), when \( x = 4 \) (which corresponds to the \( y \)-value of \( f \)), the original \( x \)-value is \( 2 \).

Step 3: Applying the Inverse Function Theorem

The inverse function theorem tells us that \( \frac{d}{dx} f^{-1}(b) = \frac{1}{f'(a)} \) where \( f(a) = b \). Here, \( a = 2 \) and \( b = 4 \) as \( f(2) = 4 \). Also, \( f'(2) = \frac{1}{3} \).

Step 4: Calculating the Derivative of the Inverse

Using the inverse function theorem formula, we substitute the known derivative: \( \frac{d}{dx} f^{-1}(4) = \frac{1}{f'(2)} = \frac{1}{\frac{1}{3}} = 3 \). The derivative of the inverse function at \( x=4 \) is \( 3 \).

Key Concepts

Differentiable Functions

Differentiable functions are a central concept in calculus, ensuring smoothness and the existence of a tangent line at every point in their domain. If a function is differentiable, it means that at every point, the change in the function can be described by a derivative, basically giving us the slope of the tangent line there.
This notion of differentiability is crucial for understanding fluctuations in graphs as it dictates where and how quickly these changes occur. For example, the function in our exercise, denoted as \( y = f(x) \), smoothly travels through the point (2,4) with a slope of \( \frac{1}{3} \).
Here, differentiability indicates that near \( x=2 \), the graph increases at a consistent rate dictated by this slope. Integrating these ideas allows us to predict or interpret graph behaviors around any given point.

Inverse Functions

Inverse functions essentially 'undo' the action of a function. Imagine if a function took input \( a \) and gave output \( b \), its inverse would find the original input value \( a \) when given \( b \).
For our exercise, the function \( f(x) \) has an inverse. This means there is a function \( f^{-1}(x) \) that can reverse the process. Given \( f(2) = 4 \), the inverse function will return \( 2 \) for the input \( 4 \).
When analyzing inverse functions, we're often interested in their differentiability and behavior. Using the inverse function theorem, we can relate the derivatives of \( f(x) \) and \( f^{-1}(x) \), especially at particular points on their graphs. The theorem helps simplify complex relationships by connecting their slopes.

Derivatives

Derivatives represent how a function changes when approaching or moving away from a particular point. Think of derivatives as the rate of change or 'speed' of a function at any given instant.
In our example, at the point (2,4), the derivative \( f'(2) \) is \( \frac{1}{3} \). This derivative tells us the slope of the tangent to the function \( f(x) \) at that precise point.
When considering the inverse function \( f^{-1}(x) \), the derivative at a corresponding point like \( x = 4 \) can be found using the inverse function theorem. According to the exercise, the derivative \( \frac{d}{dx} f^{-1}(4) \) equals \( 3 \), which is the reciprocal of the slope at \( x=2 \). This relationship is crucial in analyzing inverses as it allows us to understand how the inverse function behaves in response to changes.