Problem 41 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \text { If } f(x+y)=f(x)+f(y) \forall x, y \& f^{\prime}(1)=3 \text { , then test the differentiability of } f(x) \text { . } $$

Step-by-Step Solution

Answer

The function f(x) has the derivative f'(x) = 3 for all values of x. Since f'(x) is continuous everywhere, f(x) is differentiable at every point.

Step 1: Identify the given conditions

We have function f(x) such that: 1. f(x+y) = f(x) + f(y) for all x, y. 2. f'(1) = 3. Now, let's find the derivative f'(x).

Step 2: Differentiate f(x+y) using Chain Rule

We'll differentiate f(x+y) w.r.t x using chain rule: \(\frac{d}{dx}f(x+y) = \frac{d}{dx}(f(x) + f(y))\) First, let's differentiate the left side: \(\frac{d}{dx}f(x+y) = f'(x+y) \frac{d(x+y)}{dx} = f'(x+y)\) On the right side, \(f(x)\) will be differentiated w.r.t x and \(f(y)\) will be considered as constant as we are differentiating w.r.t x: \(\frac{d}{dx}(f(x) + f(y)) = f'(x) + 0 = f'(x)\) Now equating both the sides, we obtain: \(f'(x+y) = f'(x)\)

Step 3: Derivation using specific values of x and y

Now let's take a specific case for x and y which will help us derive the final expression for f'(x). Let \(x=1\) and \(y=x-1\), so we have: \(f'(1 + (x - 1)) = f'(x)\) However, from our given condition, f'(1) = 3: \(f'(x) = 3\)

Step 4: Differentiability test

Now, we have derived that f'(x) = 3 for all values of x. Since f'(x) is continuous everywhere in the domain, the function f(x) is differentiable at every point.

Key Concepts

Chain Rule

The Chain Rule is a fundamental tool in calculus for finding the derivative of composite functions. When we have a function that is composed of other functions, the Chain Rule allows us to differentiate it step by step.

Let's consider a composite function like \( h(x) = f(g(x)) \). The derivative of \( h(x) \) with respect to \( x \) can be found by multiplying the derivative of \( f \) at \( g(x) \) by the derivative of \( g \) at \( x \), formally written as: \[ h'(x) = f'(g(x)) \times g'(x) \].

In the given exercise, we apply the Chain Rule to the functional equation \( f(x+y) = f(x) + f(y) \). When differentiating both sides with respect to \( x \), we considered \( y \) as a constant and hence its derivative is zero. The Chain Rule simplifies the process of taking derivatives of complex expressions, revealing the underlying behavior of functions.

Derivative Test

The derivative test is a method used to determine the differentiability of a function at a given point. If a function has a derivative at every point in its domain, it's said to be differentiable on that domain. For a function \( f(x) \), if \( f'(x) \) exists and is continuous across an interval, then the function is differentiable on that interval.

Specifically, for the function in our exercise, the derivative test involves checking whether \( f'(x) \) can be found and if it remains consistent for all values of \( x \). Since we have established that \( f'(x) = 3 \) for all \( x \), the function passes the derivative test and is therefore differentiable everywhere. This means the graph of \( f(x) \) has no sharp corners or discontinuities, and it can be smoothly traced without lifting the pencil from the paper.

Functional Equation

A functional equation is an equation in which the unknowns are functions rather than simple quantities. These types of equations set relationships between the values of the function at different points. In our exercise, the functional equation is \( f(x+y) = f(x) + f(y) \), which is valid for all real values of \( x \) and \( y \).

This particular functional equation suggests that the function \( f \) has the property of additivity—that the function value of a sum is equal to the sum of the function values. This is a hallmark of linear functions, which could potentially simplify finding the derivative of such functions. It is also worth noting that the solutions to functional equations can reveal much about the nature of a function and its graph.