Problem 31 - Thomas Calculus

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

The value of \( \frac{d}{dx}f^{-1}(x) \) at \( x = -1 \) is \( \frac{1}{9} \).

Step 1: Understand the Problem

We are given a function \( f(x) = x^{3} - 3x^{2} - 1 \) and we need to find \( \frac{d}{dx}f^{-1}(x) \) at the point where \( x = f(3) = -1 \). This involves finding the derivative of the inverse function at a specific point.

Step 2: Use the Inverse Function Theorem

The Inverse Function Theorem tells us that if \( y = f(x) \) is differentiable and \( f'(x) eq 0 \), then \( \frac{d}{dy}f^{-1}(y) = \frac{1}{f'(f^{-1}(y))} \). We need to compute \( f'(x) \) and then find it at the point where \( f(x) = -1 \).

Step 3: Differentiate \( f(x) \)

Find the derivative of the function: \( f'(x) = \frac{d}{dx} (x^3 - 3x^2 - 1) = 3x^2 - 6x \).

Step 4: Find \( f'(3) \)

Substitute \( x = 3 \) into the derivative: \( f'(3) = 3(3)^2 - 6(3) = 27 - 18 = 9 \).

Step 5: Apply Inverse Function Theorem

Using the value from Step 4, \( f'(3) = 9 \), we apply the theorem: \( \frac{d}{dx}f^{-1}(-1) = \frac{1}{f'(3)} = \frac{1}{9} \).

Step 6: Conclusion

Thus, the derivative of the inverse function at the point \( x = -1 \) is \( \frac{1}{9} \).

Key Concepts

Derivative of inverse functions

The derivative of inverse functions is a vital topic in calculus and can be best understood through the Inverse Function Theorem. The idea here is that if a function has an inverse near a point, we can find the rate of change (or derivative) of this inverse function, given the original function's derivative. Consider a function \(f(x)\) and its inverse \(f^{-1}(x)\). To find \(\frac{d}{dx}f^{-1}(x)\), we apply the formula:

• \(\frac{d}{dy}f^{-1}(y) = \frac{1}{f'(f^{-1}(y))}\)

This means that the derivative of the inverse function at a particular point \(y = f(a)\) can be determined by taking the reciprocal of the derivative of the original function \(f\) evaluated at \(x = a\). It is crucial to ensure that the derivative of \(f\), \(f'(x)\), is not zero for the inverse function to be differentiable at that point.
In our exercise, we used this theorem to find \(\frac{d}{dx}f^{-1}(x)\) at a specific point by first calculating \(f'(x)\) then determining its value at the needed point of the original function.

Inverse functions

Inverse functions are central to understanding how functions can "undo" one another. If a function \( f \) takes an input \( x \) and produces an output \( y \), then the inverse function \( f^{-1} \) will take \( y \) as an input and bring us back to the original \( x \). For two functions to be true inverses, the following must be accurate:

• \(f(f^{-1}(x)) = x\)
• \(f^{-1}(f(x)) = x\)

Inverse functions are not defined for all input values. A function must be one-to-one (bijective) in the domain of interest. This ensures that for each output value in the range, there is exactly one corresponding input value in the domain. Often, graphing the function and its inverse helps visualize this relationship, where their plots are mirror images over the line \(y = x\).
In the exercise, finding the inverse was not necessary explicitly but understanding its role was crucial to apply the Inverse Function Theorem.

Differentiation

Differentiation involves finding a function's rate of change, which tells us how the function value changes as its input changes. The derivative of a function gives us this rate and can be thought of as a gradient of the function's graph. For a polynomial function \(f(x) = x^3 - 3x^2 - 1\), differentiating involves applying power rules:

• Differentiate \(x^3\) to get \(3x^2\)
• Differentiate \(-3x^2\) to get \(-6x\)

So, the derivative \(f'(x) = 3x^2 - 6x\).This derivative tells us the slope of the tangent line to the graph of \(f(x)\) at any point \(x\). For our exercise, finding \(f'(3) = 9\) was a critical step, as it enabled us to utilize the Inverse Function Theorem to find the derivative of the inverse function at a specified point.