If for all x, y the function f is defined by f(x)+f(y)+ f(x) ·f(y)=1 and f(x)>0, then (A) f^'(x)

step 1 In this problem we have fx plus fy plus fx into fy is equals to one. Let us substitute x as 0 an

If for all x, y the function f is defined by f(x)+f(y)+ f(x)...

If for all $x, y$ the function $f$ is defined by $f(x)+f(y)+$ $f(x) \cdot f(y)=1$ and $f(x)>0$, then
(A) $f^{\prime}(x)$ does not exist
(B) $f^{\prime}(x)=0$ for all $x$
(C) $f^{\prime}(0)<f^{\prime}(1)$
(D) None of these

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Key Concepts

Functional Equations

A functional equation is an equation where the unknowns are functions, and the equation involves the functions in various combinations. In this context, the equation f(x) + f(y) + f(x)f(y) = 1 holds for all x and y, forcing the function to have specific properties. Understanding functional equations involves techniques like substitution and symmetry to determine the general form of the function.

Constant Functions

A constant function is one that assigns the same value to every input in its domain. In this problem, by substituting particular values, we deduce that the function must be constant since the equation holds for all x and y. Recognizing that the function is constant simplifies analysis, as constant functions have predictable behavior, particularly under differentiation.

Derivative of a Constant Function

The derivative of a constant function is always zero because the rate of change of a constant is nil. Since the function in this problem is determined to be constant, its derivative is zero everywhere on its domain. This concept is fundamental in calculus and is used to quickly assess the behavior of functions that do not depend on the input variable.

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