Problem 56 - A Textbook of Differential Calculus

Question

Let \(f(x)=x+\cos x+2\) and \(g(x)\) be the inverse function of \(f(x)\), then \(g^{\prime}(3)\) equals to \(\ldots \ldots \ldots\)

Step-by-Step Solution

Answer

\( g'(3) = 1 \)

Step 1: Understand the Function and Its Inverse

We have the function \( f(x) = x + \cos x + 2 \). The inverse function \( g(x) \) satisfies \( f(g(x)) = x \). Our goal is to find the derivative of \( g \) at 3, which means we want \( g'(3) \).

Step 2: Use the Derivative Formula for Inverses

The derivative of an inverse function is given by \( g'(x) = \frac{1}{f'(g(x))} \). We need \( g(3) \) to use this formula, which means finding \( x \) such that \( f(x) = 3 \).

Step 3: Find g(3) by Solving the Equation

We need to solve the equation \( x + \cos x + 2 = 3 \). Simplifying, we get \( x + \cos x = 1 \). This will give us the value of \( x = g(3) \).

Step 4: Solve x + cos x = 1

The equation \( x + \cos x = 1 \) is transcendental but can be solved approximately: when \( x = 0 \), \( \cos 0 = 1 \) and \( 0 + 1 = 1 \), which satisfies the equation, hence \( x = 0 \). This implies \( g(3) = 0 \).

Step 5: Compute f'(x)

Find the derivative of \( f(x) = x + \cos x + 2 \). The derivative is \( f'(x) = 1 - \sin x \).

Step 6: Find f'(g(3))

Now compute \( f'(g(3)) \). We have \( g(3) = 0 \), so \( f'(g(3)) = f'(0) = 1 - \sin 0 = 1 \).

Step 7: Calculate g'(3)

Using the inverse function derivative formula: \( g'(3) = \frac{1}{f'(g(3))} = \frac{1}{1} = 1 \).

Key Concepts

Differential Calculus

Differential calculus is a subfield of calculus focused on the study of how functions change when their inputs change. At its core, this branch of mathematics is concerned with the notion of a derivative.
The derivative of a function at a certain point illustrates the rate at which the function's value changes as the input varies. It's like discerning how fast a car is moving by observing the change in its speedometer over time.
In our problem, we first encounter differential calculus when finding the derivative of the function \( f(x) = x + \cos x + 2 \). By differentiating, we evaluate \( f'(x) = 1 - \sin x \).
This expression portrays how \( f(x) \) changes as \( x \) varies. It's pivotal as it determines how quickly \( f(x) \) reacts to changes in \( x \), helping us reach subsequent calculations efficiently.

Derivative of Inverse Function

The concept of the derivative of an inverse function links closely to the idea that some functions have opposites or 'inverses' that reverse their effects. For example, if \( f \) is a function and \( g \) is its inverse, then when you apply \( g \) to the result of \( f \), you return to the original value: \( f(g(x)) = x \).
The beauty of the derivative of an inverse function lies in its formula: if \( y = f(x) \), then \( g'(y) = \frac{1}{f'(x)} \). This means that the derivative of the inverse function at a point is the reciprocal of the derivative of the original function at the corresponding point.
In the exercise question, we leverage this formula to find \( g'(3) \). Once we solve the equation \( f(x) = 3 \) and find that \( g(3) = 0 \), we compute the reciprocal of \( f'(x) \) at \( x = 0 \) to get \( g'(3) = 1 \). This elegantly reveals how the steepness of \( f(x) \) affects the inverse function \( g(x) \).

Transcendental Equation

A transcendental equation is one that cannot be solved using algebraic methods alone. This means it includes transcendental functions like trigonometric, exponential, or logarithmic functions.
Considering our exercise, the equation \( x + \cos x = 1 \) is transcendental. Why? Because it contains a cosine function, making it resistant to simple algebraic manipulations. Instead, evaluating such equations often requires numerical or graphical methods to find approximate solutions.
In our problem's context, observing that if \( x = 0 \), then \( \cos 0 = 1 \) leads us to the solution \( x + \cos x = 1 \). This highlights the power of intuitive reasoning in solving transcendental equations, allowing us to establish that the inverse function's value at 3 satisfies \( g(3) = 0 \). Understanding transcendental equations is crucial for solving complex mathematical problems that go beyond basic arithmetic operations.