The function `f:R→R` satisfies `f(x^2 ).f ′′ (x)=f ′ (x).f ′ (x^2 )` for all real x. Given that f(1)=1 and compute the value of `f ′ (1)+f ′′ (1)`.

The function f: RvecR satisfies f(x^2)dotf^(x)=f^(prime)(x)dotf^(prime)(x^2) for all real xdot Given that f(1)=1 and f^(1)=8 , then the value of f^(prime)(1)+f^(1) is 2 b. 4 c. 6 d. 8

A function f(x) satisfies the functional equation x^2f(x)+f(1-x)=2x-x^4 for all real x. f(x) must be

The function f:R rarr R defined as f(x)=(x^(2)-x+1)/(x^(2)+x+1) is

If a real valued function f(x) satisfies the equation f(x +y)=f(x)+f (y) for all x,y in R then f(x) is

A function f : R→R satisfies the equation f(x)f(y) - f(xy) = x + y ∀ x, y ∈ R and f (1)>0 , then

A function f : R -> R^+ satisfies f(x+y)= f(x) f(y) AA x in R If f'(0)=2 then f'(x)=

For x inR - {1} , the function f(x ) satisfies f(x) + 2f(1/(1-x))=x . Find f(2) .

A polynomial function f(x) satisfies the condition f(x)f(1/x)=f(x)+f(1/x) for all x inR , x!=0 . If f(3)=-26, then f(4)=

If f(x)=x^(2) is a real function, find the value of f(1).

For x in R , the functions f(x) satisfies 2f(x)+f(1-x)=x^(2) . The value of f(4) is equal to