Problem 7 - University Calculus: Early Transcendentals

Question

Let \(f(x)=x^{3}-3 x^{2}-1, x \geq 2 .\) Find the value of \(d f^{-1} / d x\) at the point \(x=-1=f(3)\).

Step-by-Step Solution

Answer

\(\frac{1}{9}\)

Step 1: Understanding the Problem

We need to find the derivative of the inverse function, \(f^{-1}(x)\), at the point where \(x = -1\). The problem specifies that \(f(3) = -1\), which implies \(f^{-1}(-1) = 3\).

Step 2: Finding the Derivative of f(x)

First, we differentiate the function \(f(x) = x^3 - 3x^2 - 1\). The derivative \(f'(x)\) is calculated as follows:\[f'(x) = \frac{d}{dx}(x^3 - 3x^2 - 1) = 3x^2 - 6x.\]

Step 3: Evaluating f'(x) at x=3

To find \(df^{-1}/dx\) at \(x = -1\), we need to evaluate \(f'(x)\) at \(x = 3\). Substituting \(x = 3\) into \(f'(x)\), we have:\[f'(3) = 3(3)^2 - 6(3) = 27 - 18 = 9.\]

Step 4: Finding the Derivative of the Inverse Function

The derivative of the inverse function \(f^{-1}(x)\) with respect to \(x\) is given by the formula:\[\frac{df^{-1}}{dx} \bigg|_{x = -1} = \frac{1}{f'(f^{-1}(-1))} = \frac{1}{f'(3)}.\]Substituting \(f'(3) = 9\), we get:\[\frac{df^{-1}}{dx} \bigg|_{x = -1} = \frac{1}{9}.\]

Key Concepts

Inverse Functions

Inverse functions are pairs of functions that reverse each other's effects. So, if you have a function, like our given function \( f(x) = x^3 - 3x^2 - 1 \), its inverse function \( f^{-1}(x) \) will satisfy the relationship \( f(f^{-1}(x)) = x \). This means that applying the function and then its inverse will bring you back to your original input. Understanding inverse functions is crucial because they allow us to determine inputs from known outputs. In practical applications, they help solve equations where we need to find what value of \( x \) gives a particular result. One thing to note is that not all functions have inverses. A function must be one-to-one (bijective) for its inverse to exist. In the case of our function \( f(x) \), restrictions like \( x \geq 2 \) can ensure it is a one-to-one function, making its inverse valid over this domain. Remember, finding an inverse function can involve swapping the roles of \( x \) and \( y \) (interchanging inputs and outputs) in a function's equation.

Differentiation

Differentiation is a process in calculus used to calculate the rate at which a function is changing at any point. Differentiating a function gives us what we call the derivative, denoted \( f'(x) \), which represents this rate of change. In our problem, we start with the function \( f(x) = x^3 - 3x^2 - 1 \). Differentiating this function involves finding \( f'(x) \). By applying the power rule, which says the derivative of \( x^n \) is \( nx^{n-1} \), we compute:

• For \( x^3 \), the derivative is \( 3x^2 \).
• For \(-3x^2\), the derivative is \(-6x\).
• The constant \(-1\) becomes 0.

Thus, the derivative is \( f'(x) = 3x^2 - 6x \).Using this derivative, we can discover how the original function changes as \( x \) changes, which is pivotal in solving many calculus problems. It also plays a key role in finding the derivative of the inverse function, as seen in this exercise.

Calculus Problems

Solving calculus problems often involves multiple steps, integrating various concepts such as differentiation and inverse functions. This particular exercise is an excellent illustration of these concepts working together. First, understand the problem's requirements. Here, we need the derivative of an inverse function at a specific point. We achieve this by:

• Finding the function's derivative \( f'(x) \).
• Evaluating \( f'(x) \) at the point related to our interest, which is \( x = 3 \) derived from \( f(3) = -1 \).
• Using the formula for the derivative of the inverse function: \( \frac{df^{-1}}{dx} = \frac{1}{f'(f^{-1}(-1))} \).

Solving this yields \( \frac{1}{9} \).Such exercises enhance your understanding of how functions behave and change, equipping you with tools to tackle even more complex calculus challenges. By practicing these problems, you develop problem-solving skills and a deeper insight into mathematical relationships.