Problem 13 - Calculus 1 Course in Mathematics for the IIT JEE and Other Engineering Exams

Question

Given \(f(x+y+z)=f(x) f(y) f(z)\) for all \(x, y, z\) If \(f(2)=4\) and \(f^{\prime}(0)=3\), then \(f^{\prime}(2)=\) (a) 12 (b) \(-12\) (c) \(+12\) or \(-12\) (d) 1

Step-by-Step Solution

Answer

The correct answer is \(f'(2) = 12\).

Step 1: Understanding the Functional Equation

Given the function satisfies: \(f(x+y+z) = f(x) f(y) f(z)\) for all \(x, y, z\). This suggests a form reminiscent of an exponential function since it repeats across addition of three variables.

Step 2: Postulating a Solution

Given the nature of the functional equation, suppose \(f(x) = e^{g(x)}\). This turns the multiplication of function values into an addition form: \(e^{g(x+y+z)} = e^{g(x) + g(y) + g(z)}\), which implies \(g(x+y+z) = g(x) + g(y) + g(z)\). This is a form of Cauchy's functional equation.

Step 3: Using Initial Conditions

Given \(f(2) = 4\), and using our postulate \(f(x) = e^{g(x)}\), it follows \(e^{g(2)} = 4 \implies g(2) = \ln 4\).

Step 4: Differentiating the Function

Differentiating \(f(x) = e^{g(x)}\), we have \(f'(x) = e^{g(x)} g'(x)\). We know \( f'(0) = 3 \), and \( f(0) = e^{g(0)} = 1 \), hence \(g'(0) = 3\) because \(f(0) = e^{g(0)} = 1\) by exponential uniqueness, \(g(0) = 0\).

Step 5: Solving for Derivative at Specific Point

We use \(f'(x) = e^{g(x)} g'(x)\) and find \(f'(2) = e^{g(2)} g'(2) = 4g'(2)\). Since \(g(x)\) is linear from the derivation equation \(g(x+y+z) = g(x) + g(y) + g(z)\) and is differentiable, \(g(x) = cx\) so \(g'(x) = c\) and \(g'(0) = c = 3\). Thus, \(g'(2) = 3\).

Step 6: Calculating the Final Result

Plug back the constant value, find \(f'(2) = 4g'(2) = 4 \cdot 3 = 12\).

Key Concepts

Exponential Functions

Exponential functions are fundamental in mathematics and play a significant role in various fields like calculus, physics, and economics. An exponential function can be expressed in the form \( f(x) = e^{g(x)} \), where \( e \) is the base of the natural logarithm, approximately equal to 2.71828.
Exponential functions exhibit unique properties like an initial value set at \( x=0 \), known as the y-intercept which, for exponential functions, is usually \( f(0) = 1 \) if no transformations are applied.

• Growth or decay is characterized by the exponent \( g(x) \).
• If \( g(x) \) is linear, the function grows or decays at a constant rate.
• Their graphs are non-linear, typically J-shaped, representing exponential growth or an inverted J representing decay.

In the context of functional equations like the one given in the problem, the structure \( f(x+y+z) = f(x)f(y)f(z) \) is suggestive of multiplication of probabilities or repeated compounded growth, hence the connection to an exponential format.

Differential Calculus

Differential calculus is a branch of mathematics that focuses on how functions change when their inputs change. The key idea is differentiation, which finds the derivative—a measure of how a function changes as its input changes.
The derivative of a function \( f(x) \) with respect to \( x \), denoted \( f'(x) \), describes the rate at which \( f(x) \) is changing at any particular point.

• If the derivative is positive, the function is increasing.
• If it is negative, the function is decreasing.
• When the derivative is zero, the function is constant or is at a local maximum/minimum.

When evaluating specific points like \( f'(0) = 3 \), it implies that at \( x = 0 \), the function \( f(x) \) is changing at a rate of 3 units per unit change in \( x \).

The Role of Differentiation in the Exercise

In our exercise, we differentiated the function \( f(x) \) hypothesized as \( e^{g(x)} \). We then used the condition \( f'(0) = 3 \) to derive a constant rate of change for \( g(x) \), and subsequently confirmed that \( f'(2) = 12 \) using this derivative format.

Cauchy's Functional Equation

Cauchy's functional equation is an equation of the form \( g(x + y) = g(x) + g(y) \). This equation is critical in understanding linearity within functions.
The general solution to Cauchy's equation, when \( g \) is continuous, is \( g(x) = cx \) for some constant \( c \). This linear form provides the basis for many results in functional equations.
In this exercise, this form was extended to \( g(x + y + z) = g(x) + g(y) + g(z) \), reinforcing that the underlying function \( g \) is indeed linear provided the function is differentiable everywhere.

Application of Cauchy's Equation in the Exercise

By assuming the structure \( f(x) = e^{g(x)} \), and recognizing that \( f(x+y+z) = f(x)f(y)f(z) \) leads to the expanded equation \( g(x+y+z) = g(x) + g(y) + g(z) \), we see the application of Cauchy’s equation in deducing that \( g(x) \) must be linear and continuous, hence \( g(x) = 3x \) given \( g'(0) = 3 \). This foundation helped us determine the value of \( f'(2) \) accurately.