If f(x+y+z)=f(x) ·f(y) ·f(z) for all x, y, z in R such that f(2)=4, f^'(0)=3, find f^'(2).

Name: Problem 92, Chapter 4: Differential Calculus Booster with Problems and Solutions for JEE Main and Advanced
Uploaded: 2022-03-14T18:09:49-08:00
Duration: 4 min 15 s
Channel: Gaurav Kumar
Description: If f(x+y+z)=f(x) ·f(y) ·f(z) for all x, y, z in R such that f(2)=4, f^'(0)=3, find f^'(2).

step 1 Hello everyone we are going to solve a question in this question we are given a function f of x

If f(x+y+z)=f(x) ·f(y) ·f(z) for all x, y, z in R such that...

Key Concepts

Functional Equation

A functional equation is an equation where the unknowns are functions. In this context, the equation f(x+y+z) = f(x) · f(y) · f(z) is used to derive properties of f, especially by choosing particular values for the variables (such as zero) to reduce the equation to more standard forms.

Exponential Functions

Functions that satisfy equations like f(x+y) = f(x) · f(y) are typically exponential functions. With additional regularity conditions, one is led to conclude that f has the form f(x) = e^(kx) (or another equivalent exponential form), where the constant k is determined by given initial conditions.

Differentiation of Functional Equations

Differentiating a functional equation with respect to one of its parameters helps relate the function's derivative at different points. For an exponential function f(x) = e^(kx), one finds that f'(x) = k · e^(kx) = k · f(x), so the derivative at any point is directly proportional to the function's value at that point.

Initial Conditions and Parameter Determination

Initial or boundary conditions, such as f(2) = 4 and f'(0) = 3, are used to determine the unknown parameters in the general solution of the functional equation. Once the constants are found, one can compute other quantities like f'(2) by using the relationship between f and f'.

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