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Chapter 3: Problem 6

Assume that \(f: \mathbb{R} \rightarrow \mathbb{R}\) is invertible and differentiable. Compute \(\left(f^{-1}\right)^{\prime}(4)\) from the given information. $$ f^{-1}(4)=1, f^{\prime}(s)=4+\pi \cos (\pi \mathrm{s}) $$

Short Answer

Expert verified
The value of \((f^{-1})'(4)\) is \(\frac{1}{4 - \pi}\).

Step by step solution

01

Interpret the Given Information

We are provided that \(f^{-1}(4) = 1\), which means that \(f(1) = 4\). The derivative of \(f\) is given as \(f'(s) = 4 + \pi \cos(\pi s)\). We aim to find \(\left(f^{-1}\right)'(4)\).
02

Use the Derivative of the Inverse Function Theorem

The theorem states that if \(f\) is differentiable and its inverse \(f^{-1}\) exists at a point \(f(a)\), then the derivative of the inverse at this point is given by:\[\left(f^{-1}\right)'(f(a)) = \frac{1}{f'(a)} \]Since we have \(f^{-1}(4) = 1\), it follows that \(f(a) = 4\) where \(a = 1\). Therefore, we need \(f'(1)\) to compute \(\left(f^{-1}\right)'(4)\).
03

Compute \(f'(1)\)

Substitute \(s = 1\) into the derivative expression for \(f\):\[f'(s) = 4 + \pi \cos(\pi s)\]\[f'(1) = 4 + \pi \cos(\pi \times 1) = 4 + \pi \cos(\pi)\]Calculate the cosine term: \(\cos(\pi) = -1\), therefore\[f'(1) = 4 + \pi (-1) = 4 - \pi\]
04

Calculate \(\left(f^{-1}\right)'(4)\)

Using the result from Step 2,\[\left(f^{-1}\right)'(4) = \frac{1}{f'(1)} = \frac{1}{4 - \pi}\]
05

Final Answer

The computed derivative of the inverse function at the given point is\[\left(f^{-1}\right)'(4) = \frac{1}{4 - \pi}\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiable Functions
Differentiable functions play a pivotal role in calculus. They are functions that have a derivative at each point in their domain. This derivative is a function that provides the slope of the tangent to the curve at each point.
  • A function is said to be differentiable if its derivative exists at every point in the domain.
  • For a real-valued function of a real variable, say \(f(x)\), the derivative is denoted as \(f'(x)\).
  • The existence of the derivative implies continuity, meaning the function doesn't have any abrupt jumps or breaks.
Differentiability is essential for finding inverse functions, as it ensures that the original function behaves smoothly and predictably. When a function is invertible and differentiable, like in our exercise with the function \(f\), you can apply specific theorems to find the derivative of the inverse.
Inverse Function Derivative
The Inverse Function Theorem is a key concept that helps us compute the derivative of an inverse function if it exists. When a function is invertible and differentiable, the theorem offers a way to calculate the derivative of the inverse.
  • For a function \(f\) with an inverse \(f^{-1}\), if \(f\) is differentiable at \(a\) and its derivative at \(a\), \(f'(a)\), is not zero, then the inverse function \(f^{-1}\) is differentiable at \(f(a)\).
  • The derivative of the inverse function at point \(f(a)\) is given by \(\left(f^{-1}\right)'(f(a)) = \frac{1}{f'(a)}\).
In our problem, we are tasked to compute \(\left(f^{-1}\right)'(4)\). Knowing that \(f(1) = 4\) and \(f'(1)\) is calculated as \(4 - \pi\), the application of the theorem allows us to conclude that \(\left(f^{-1}\right)'(4) = \frac{1}{4 - \pi}\). This shows the power of the theorem in providing a straightforward yet elegant solution.
Trigonometric Functions
Trigonometric functions, such as sine and cosine, are fundamental in math, appearing frequently in calculus problems, including derivatives. In our exercise, the derivative of \(f\) involves a trigonometric component: \(4 + \pi \cos(\pi s)\).
  • Cosine, denoted as \(\cos(s)\), measures the adjacent side over hypotenuse in a right triangle for an angle \(s\).
  • It is a periodic function, meaning its values repeat every \(2\pi\) radians.
  • When differentiating functions involving trigonometric parts, recognizing identities and periodic properties of these functions is helpful. For instance, \(\cos(\pi) = -1\).
The periodic nature of cosine plays a crucial role in calculating the function's behavior at specific points. When \(s = 1\), \(\cos(\pi \times 1) = -1\) simplifies our calculations, allowing us to determine that \(f'(1) = 4 - \pi\). This pivotal step in computation further highlights how trigonometric functions are seamlessly integrated into solving calculus problems.

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